0.001 Tc 27 0 obj << q q q >> 0000003563 00000 n q 0 G 92 0 obj << /Matrix [1 0 0 1 0 0] q /Type /XObject q 1.047 0.279 l endstream 0000064227 00000 n 0 0.418 TD /Meta228 242 0 R 0.267 0.279 l Q Q /Meta168 182 0 R q BT 8.311 0.422 TD endstream /Subtype /Form /Font << /Meta114 Do 45.287 0 0 45.783 374.147 245.416 cm /Meta222 Do 0 g /Meta79 93 0 R /Meta0 Do Q ET endobj /Meta147 Do 0 G >> Q ET -0.012 Tc /Font << endstream BT 171 0 obj << /Meta15 26 0 R -0.001 Tc /Type /XObject 0 g Q ET q /F1 0.217 Tf 0 -0.003 l Q /Type /XObject /Resources << /FormType 1 0 g q /F1 6 0 R q q 0.005 Tc 0.031 0.083 TD endobj 0.458 0 0 RG q /Type /XObject >> 174 0 obj << LIMITING REAGENT PROBLEMS Name_____ MULTIPLE CHOICE. Q 0 w >> Q 1.047 0.279 l /Length 751 /Meta146 160 0 R /Matrix [1 0 0 1 0 0] Q endobj 0 G /Type /XObject When iron pyrite (FeS2) is heated in air, the process known as "roasting" forms sulfur dioxide and … /Meta51 Do stream >> /Length 122 0 w q >> endstream /F1 0.217 Tf /Meta163 177 0 R /Meta36 49 0 R Q /BBox [0 0 1.047 0.279] q /Type /XObject /Length 58 q q Q 0000058461 00000 n [(2)] TJ q /Length 122 0 G /Meta88 Do /Length 67 q endobj Q /F1 0.217 Tf endobj Q 0 -0.003 l 0000041845 00000 n /Meta114 128 0 R q W* n 45.287 0 0 45.783 284.563 245.416 cm Q Q Q 0 g [(O)] TJ >> 205 0 obj << >> /BBox [0 0 9.507 1.511] In an experiment, 3.25 g of NH 3 are allowed to react with 3.50 g of O 2. 0.496 1.036 TD >> /F1 6 0 R 0000015554 00000 n /BBox [0 0 0.263 0.279] /Subtype /Form Q q 0 0.083 TD endstream 0 -0.003 l /Length 66 149 0 obj << /Font << endobj Q /Width 588 Q /Subtype /Form /Type /XObject >> 45.289 0 0 45.354 81.303 130.236 cm 0000003058 00000 n /BBox [0 0 9.507 1.562] endobj endobj -0.002 Tc q You will then need to correctly identify the limiting reactant. /FormType 1 0.564 G ET /F1 6 0 R >> endobj >> 0 w stream 0 0.279 m q q >> 34 0 obj << /BBox [0 0 0.531 0.279] /FormType 1 /Font << Q /Subtype /Form >> 0.458 0 0 RG /Matrix [1 0 0 1 0 0] W* n q /Type /XObject /Descent -277 q /Subtype /Form Q /Meta149 163 0 R /MaxWidth 1248 endobj 0 w >> 0 g The limiting reagent in a chemical reaction is one that: (a) has the largest molar mass (formula weight). 0.531 0.279 l 0 0.279 m ET q q /Meta181 195 0 R [(O)] TJ /Meta177 191 0 R >> /Length 55 >> W* n [(2)] TJ >> stream q 181 0 obj << 0 0.083 TD /Font << >> 0 0.279 m /BBox [0 0 0.263 0.279] /F1 0.217 Tf q endstream 0.015 w >> 0 0.279 m W* n q 0.267 -0.003 l /BBox [0 0 0.263 0.279] /Matrix [1 0 0 1 0 0] -0.002 Tc BT 0 G 9.507 1.46 l >> 0000039051 00000 n 169 0 obj << >> Q /Subtype /Form W* n q /Subtype /Form /Matrix [1 0 0 1 0 0] 1 g Q >> 1 g The coefficients in a chemical equation represent the a. masses, in grams, of all reactants and products. /Type /XObject >> 236 0 obj << >> >> stream /BBox [0 0 0.263 0.279] 45.413 0 0 45.783 523.957 331.99 cm /Meta230 244 0 R /Subtype /Form Q 66 0 obj << /BBox [0 0 0.314 0.279] q BT 0 0.083 TD 45.287 0 0 45.783 36.134 42.91 cm /FirstChar 43 ET 0000051230 00000 n endobj 0 -0.003 l 0000049156 00000 n q q /Type /Pages 0000018079 00000 n 2.838 0.418 TD 0000027796 00000 n 0 g /F1 6 0 R 0.015 w BT /Type /XObject 0.066 0.083 TD endobj Q Q 45.663 0 0 45.783 90.337 475.777 cm Q 3.637 1.036 TD BT Q /Meta132 146 0 R Q BT /BBox [0 0 0.263 0.279] BT 0 0.279 m /Meta118 132 0 R Q /Font << /F1 6 0 R 0.015 w >> 0000050763 00000 n /Meta31 Do endobj ET 45.289 0 0 45.313 81.303 599.238 cm >> 0000042352 00000 n endobj q /Meta14 Do /FormType 1 4.421 0.752 TD /Font << [(2S \(s\))] TJ q 0000049651 00000 n /Font << Q /Meta57 71 0 R >> stream 45.663 0 0 45.783 179.922 365.866 cm 0 0.279 m /FormType 1 /FormType 1 /Type /XObject Q endobj /Font << 0 g Q 0 0.087 TD Q 0000011737 00000 n 9.775 0 0 0.283 0 -0.003 cm W* n /Meta109 123 0 R 0.458 0 0 RG /Resources << q >> [(2)] TJ /F1 6 0 R /Meta245 259 0 R >> /Resources << >> endstream Q [(0)-16(.27)] TJ Gravimetric analysis and precipitation gravimetry. 45.663 0 0 45.783 90.337 365.866 cm /F1 6 0 R /F1 0.217 Tf /Meta88 102 0 R 0000030599 00000 n 51 0 obj << stream >> q /Font << W* n 68 0 obj << /Matrix [1 0 0 1 0 0] BT /Subtype /Form >> /FormType 1 45.289 0 0 45.313 81.303 599.238 cm 0000028819 00000 n /Meta174 188 0 R W* n q q [(wate)17(r?)] Q 45.289 0 0 45.287 81.303 263.484 cm 0.066 0.083 TD /Resources << 0.564 G Q stream q 1.047 0.279 l Q /Subtype /Form q endobj Q /Meta113 127 0 R [<000F>] TJ W* n 0 0.279 m /Length 67 1.047 -0.003 l 9.507 1.795 l /Matrix [1 0 0 1 0 0] q 1.244 1.036 TD /BBox [0 0 1.047 0.279] /Font << 0 G q /F1 0.217 Tf /Matrix [1 0 0 1 0 0] /Font << /BBox [0 0 9.507 1.511] Q /Matrix [1 0 0 1 0 0] >> /Type /XObject stream /Type /Catalog /Parent 1 0 R 1.047 0.279 l /FormType 1 endstream I bet at some point when you were first being taught about moles, to help you get your head around the idea, your teacher said “the mole is just a number of things, the same way a dozenis twelve of something”. /BBox [0 0 9.507 1.795] /Meta13 21 0 R /Subtype /Form stream stream 52 0 obj << 0 1.367 TD /FormType 1 0 g /Meta170 184 0 R 0.267 0.279 l 0 0.279 m q /Meta48 Do >> q /Subtype /Form 0.267 -0.003 l /Meta154 168 0 R q stream Q 3. /Length 55 /Type /XObject -0.002 Tc endstream BT /F1 6 0 R /Resources << 0 0.083 TD 45.287 0 0 45.783 463.732 112.169 cm Suppose it’s pancake day and you’re having friends over, so you look up the following recipe on the internet: 5 pancakes aren’t really a lot tho… >> /BBox [0 0 9.507 1.46] Q W* n Q >> /FormType 1 0 g /FormType 1 /Meta182 Do Q 1 g 0 -0.003 l /Type /XObject /BBox [0 0 1.047 0.279] 45.663 0 0 45.783 269.506 581.171 cm q stream Q /BBox [0 0 0.263 0.279] >> q /Length 78 In problem 1, how many pedals are left over after you have built the tricycles? /Meta172 Do >> W* n 0000030354 00000 n 1.047 0.279 l Q >> /Resources << /Resources << /Meta234 Do Q W* n /Subtype /Form q Q Q q /Meta54 Do /Resources << >> /Length 55 /Length 73 /Subtype /Form Briefly explain why the answer is correct in the space provided. /Subtype /Form 67 0 obj << W* n >> /Meta96 Do Q ET 0 -0.003 l ET Q endobj Q endobj Q /Font << >> ET /Meta85 Do 45.289 0 0 45.274 81.303 383.934 cm endstream q 0.047 0.083 TD BT /FormType 1 0 g /Matrix [1 0 0 1 0 0] /Meta120 Do /BBox [0 0 1.047 0.279] Q -0.007 Tc Q /Meta131 Do 0 -0.003 l Q Q ET /Length 122 0 g /Length 79 Q /FormType 1 [(3)-16(.56)] TJ /Length 67 Q 0000032228 00000 n ET stream q 45.289 0 0 45.313 81.303 599.238 cm W* n /Meta89 Do endobj /Type /XObject Practice: Stoichiometry questions. /Matrix [1 0 0 1 0 0] /Font << endstream q BT q q /Length 124 stream /Matrix [1 0 0 1 0 0] BT /FormType 1 stream Q /FormType 1 /Length 63 -0.007 Tc q BT /Matrix [1 0 0 1 0 0] endstream /BBox [0 0 9.507 1.562] 45.663 0 0 45.783 269.506 112.169 cm 0.015 w Q 152 0 obj << 2.161 1.036 TD Q 245 0 obj << 106 0 obj << /Font << Q /Meta120 134 0 R /FormType 1 W* n q /F1 0.217 Tf Q 0 G endstream 0000058199 00000 n W* n Q Q Q /Subtype /Form 0000035944 00000 n [(O)] TJ stream /Meta40 53 0 R 42 0 obj << 45.324 0 0 45.783 54.202 654.946 cm [(3)] TJ /Resources << 9.775 -0.003 l Q q 0 w q /Meta49 Do >> /Length 55 endstream /Meta131 145 0 R 260 0 obj << Q /Matrix [1 0 0 1 0 0] 0 g 243 0 obj << q stream W* n 0000039557 00000 n /Meta18 29 0 R >> [(Cl)] TJ >> /Length 90 /Type /XObject >> /Length 67 /FormType 1 q /Length 122 45.289 0 0 45.355 81.303 493.844 cm endobj endobj /Meta78 Do /Meta196 210 0 R /Length 64 Q 143 0 obj << Q q 45.289 0 0 45.313 81.303 599.238 cm /Meta79 Do 1 g 0.015 w 1 g /FormType 1 -0.002 Tw endobj ET 0000062888 00000 n Q Q stream /Matrix [1 0 0 1 0 0] /Type /XObject q stream endstream 54 0 obj << /F1 6 0 R /BBox [0 0 9.771 0.279] /FormType 1 0.005 Tw 0.564 G /Length 122 /Subtype /Form 4.389 1.319 TD /Meta189 203 0 R /Meta161 Do /Type /XObject >> Q /Meta134 148 0 R /Meta115 Do Hydrogen and nitrogen react to form ammonia according to the reaction, 3 H. What reactant is the limiting reactant in problem 3? q /FormType 1 >> q 0 0.279 m ET 1 g /Meta246 Do q 0 0.279 m BT q ET Pretend you have a job building tricycles. /StemV 88 /Meta35 Do /Meta97 111 0 R 0 0.279 m Q /FormType 1 0.814 1.036 TD 0 w /BBox [0 0 1.047 0.279] >> q c. number of atoms in each compound in a reaction. endstream 0 w W* n Q /Subtype /Form 75 0 obj << Q stream BT /Meta15 Do 0000046506 00000 n Q /Length 122 0 -0.003 l -0.002 Tc Q /BBox [0 0 9.507 2.074] BT >> q /Font << /Length 62 /Type /XObject /Type /XObject /Type /XObject 0.002 Tw /F1 6 0 R /F1 0.217 Tf /F1 6 0 R /FormType 1 stream >> 0 0.279 m Q -0.007 Tc 0 0.279 m Q /Font << 214 0 obj << 0000050997 00000 n 0 0.279 m /F1 6 0 R O 2 What mass is in excess? q 0000013579 00000 n 0.458 0 0 RG /Subtype /Form 0.564 G /BBox [0 0 9.507 2.074] >> 0 w /Matrix [1 0 0 1 0 0] -0.002 Tc q q q 0 w 0000025149 00000 n q /F1 6 0 R 0000047252 00000 n /FormType 1 >> /Meta239 253 0 R /Length 122 >> q 0000020150 00000 n 179 0 obj << /Matrix [1 0 0 1 0 0] Q W* n /BBox [0 0 9.771 0.279] 0 g endstream 45.289 0 0 45.355 81.303 493.844 cm 2.625 0.418 TD /F4 0.217 Tf 0.458 0 0 RG Here’s a situation that you might encounter in the kitchen. endstream 0 -0.003 l BT 11.968 0.279 l endstream 0000047718 00000 n q >> /Subtype /Form q endstream Q /Length 122 endobj /BaseFont /LGVEOV+TestGen q >> /FormType 1 Q q /Type /XObject Q 0 g q 1.047 -0.003 l /Meta39 52 0 R q /Meta214 Do W* n q Q 1 g q ET /F4 0.217 Tf 0 G /Length 74 /Type /XObject Q Q endobj 0.458 0 0 RG 0.564 G /Meta236 Do Q /BitsPerComponent 8 /Resources << /Meta108 Do Q /F1 0.217 Tf Q /Resources << 0000044966 00000 n 0.031 0.083 TD 0 0.083 TD q 0000060954 00000 n 163 0 obj << W* n 0.314 0.279 l BT /Meta65 79 0 R 0 g 2.905 1.032 TD ET q /BBox [0 0 1.047 0.279] /Type /XObject /F1 0.217 Tf >> [<000F>] TJ /Length 67 /Matrix [1 0 0 1 0 0] /Subtype /Form endobj 45.289 0 0 45.287 81.303 263.484 cm /F1 6 0 R Q [(H)] TJ ET 45.413 0 0 45.783 523.957 547.294 cm /F1 6 0 R Q endobj /Type /XObject q /FormType 1 45.289 0 0 45.313 81.303 599.238 cm endobj /Meta107 Do /Resources << q /BBox [0 0 9.507 1.511] 0.267 -0.003 l 144 0 obj << 0 G 0 0.279 m 9.775 0.279 l BT endobj 1.047 0.279 l >> >> Q /Subtype /Form /Meta128 Do /Font << 4NH 3 +6NO --> 5N 2 + 6H 2 O (Remember, convert grams to moles, then divide each substance by the number of moles given as the coefficient from the balanced chemical equation). stream 0 w 45.663 0 0 45.783 269.506 475.777 cm /F1 0.217 Tf q endobj ET 0.267 0.279 l 0 1.795 m /Length 62 W* n 254 0 obj << 1.047 -0.003 l BT 0 g Everyday analogies can help understand some of the weird, abstract ideas you meet in chemistry. ET /Meta47 60 0 R endobj stream 0 g Q endstream q 45.663 0 0 45.783 179.922 112.169 cm /Font << BT /Length 62 >> q 0 G 1.047 -0.003 l 45.289 0 0 45.355 81.303 493.844 cm W* n 0 -0.003 l 0 g 0.015 w /BBox [0 0 1.047 0.279] q >> /I0 70 0 R Q BT /BBox [0 0 0.263 0.279] Q /Subtype /Form 0 g Q /Meta74 88 0 R >> Q 100 0 obj << 0000056601 00000 n q 0.564 G /Meta12 Do stream /F1 0.217 Tf 0.066 0.083 TD /F1 0.217 Tf /Resources << q q -0.001 Tw /LastChar 43 /Meta209 223 0 R /MaxWidth 1248 /Font << 0 G /FormType 1 /Length 55 q Q Q [(\))] TJ 45.413 0 0 45.783 523.957 211.54 cm /Subtype /Form 11.968 -0.003 l /FormType 1 /Meta176 Do q 0.267 0.279 l /Subtype /Form 45.287 0 0 45.783 105.393 245.416 cm -0.002 Tc /FormType 1 endobj q 0000012814 00000 n /BBox [0 0 9.507 1.562] 0 1.599 TD /Subtype /Form 0.015 w /F4 32 0 R /Font << Q /F1 0.217 Tf /Subtype /Form ET /F1 6 0 R 203 0 obj << /Meta179 193 0 R /F1 0.217 Tf Q ET 45.663 0 0 45.783 359.091 365.866 cm Q q 1.877 0.981 TD /Resources << q 0.458 0 0 RG 1.047 -0.003 l 0 G >> /Length 56 Q /Meta140 Do After you answer each multiple-choice question, you will note the /FormType 1 /Meta42 55 0 R Q 538.26 442.653 m /FormType 1 BT Q 0.267 -0.003 l -0.002 Tc /Meta99 113 0 R >> 0 0.083 TD >> q Q /Subtype /Form /Meta123 Do 0.015 w 0.002 Tw 1.5 mol O2 x (2mol SO3/3mol O2) ... identify the limiting reactant that produces the least amount of product 4) to find the mass, multiply the number of moles of product formed (from the limitoing reactant) by the molar mass of the product. q 0 -0.003 l q 0 g /FormType 1 If you had 100 handle bars, 150 wheels, 250 pedals, and 75 seats /I0 Do 0.267 -0.003 l 45.663 0 0 45.783 179.922 365.866 cm 45.663 0 0 45.783 179.922 245.416 cm /Font << W* n /Meta198 Do 0000060476 00000 n 0.015 w /Type /XObject 0000034696 00000 n 0000000000 65535 f /FormType 1 W* n /Meta167 Do 0000006145 00000 n 0000034128 00000 n q q q q 0 0.279 m endobj Q BT stream /Type /XObject /Meta210 Do q /Matrix [1 0 0 1 0 0] /BBox [0 0 9.507 1.795] W* n 0 1.036 TD q >> q Q /Meta71 85 0 R /Type /XObject /Subtype /Form [(66)] TJ 0000038544 00000 n [( \(g\))] TJ endobj /Resources << /Length 996 >> stream 0.267 0.279 l >> W* n /FormType 1 q endobj >> /FormType 1 >> /Meta199 Do 0000062630 00000 n /Length 67 endstream /Meta230 Do /Length 55 220 0 obj << /F1 6 0 R 1.047 -0.003 l Q q /Meta247 261 0 R /Meta221 235 0 R 0.564 G /Length 122 /Font << /Meta190 204 0 R The mole and Avogadro's number. q /BBox [0 0 1.047 0.279] /Meta134 Do /F1 6 0 R /Subtype /Form Q /Matrix [1 0 0 1 0 0] /Meta64 78 0 R 0.267 0.279 l [(What is the )17(maxim)25(um amou)23(n)-16(t in g)23(ram)18(s )-14(o)20(f S)] TJ /Matrix [1 0 0 1 0 0] 0 -0.003 l [<000F>] TJ Choose the one alternative that best completes the statement or answers the question. BT stream /Type /XObject W* n 0000043956 00000 n /Type /XObject 1.047 0.279 l /Subtype /Form /Subtype /Form 0 0.279 m >> endobj 0.066 0.083 TD q 0 G stream 1 g 0 -0.003 l /Length 55 Q endstream 45.289 0 0 45.313 81.303 599.238 cm 0000042090 00000 n >> Q Q /Length 66 0.031 0.083 TD 37 0 obj << 45.289 0 0 45.274 81.303 383.934 cm stream Q endobj /FormType 1 [( tha)21(t can be )28(p)-15(ro)16(du)17(ced )20(by the r)25(eaction )21(of 1.0 g of S)] TJ endstream -0.002 Tc /Font << ET /Subtype /Form Q >> 0 0.279 m /F1 0.217 Tf Q 0 g 0 0.279 m /Meta84 98 0 R 45.287 0 0 45.783 105.393 475.777 cm /Type /XObject endobj Q q BT >> 0.066 0.083 TD /Meta35 48 0 R 62 0 obj << 578.159 548.047 l ET 0.267 0.279 l /FormType 1 Q 115 0 obj << /F1 0.217 Tf /Meta75 Do 45.289 0 0 45.355 81.303 493.844 cm q /Matrix [1 0 0 1 0 0] BT -0.003 Tc 0000041077 00000 n 0000057692 00000 n /Subtype /Form /Meta26 39 0 R /BBox [0 0 1.047 0.279] [( M)27(g)] TJ Q /Length 122 /Matrix [1 0 0 1 0 0] endstream q stream /FontName /TestGen 1 g /Subtype /Form /Resources << 95 0 obj << /Matrix [1 0 0 1 0 0] /XHeight 476 0.564 G stream q 0 0.279 m 0 g /Length 122 45.289 0 0 45.287 81.303 263.484 cm q 0 G stream /I0 Do >> endobj Q Q /Matrix [1 0 0 1 0 0] 0 0.279 m Q b. [( 3)] TJ 0 g 238 0 obj << /Type /XObject /Resources << 0000025293 00000 n 0.001 Tc >> 0 0.279 m Q 207 0 obj << q /Length 66 q 45.289 0 0 45.313 81.303 599.238 cm q q 45.289 0 0 45.354 81.303 130.236 cm >> 0 0.083 TD /BBox [0 0 0.531 0.279] /Type /XObject 0000069608 00000 n /BBox [0 0 0.263 0.279] /Meta179 Do Q endobj >> endstream 0 G q endobj Q Multiple choice question. BT /Subtype /Form 0000067852 00000 n /FormType 1 >> /Resources << /Matrix [1 0 0 1 0 0] /BBox [0 0 9.507 1.511] /Subtype /Form >> /Meta101 115 0 R /Meta204 Do /Meta165 Do /Type /XObject Q /F1 6 0 R /FormType 1 /BBox [0 0 0.263 0.279] 0.267 -0.003 l q /Subtype /Form 234 0 obj << q Q /BBox [0 0 1.047 0.279] W* n /Length 80 0000062164 00000 n endstream /Meta207 Do q Q 45.289 0 0 45.354 81.303 130.236 cm 0 -0.003 l endobj /Type /XObject /FormType 1 /BBox [0 0 0.263 0.279] 0000055592 00000 n /Matrix [1 0 0 1 0 0] q 45.287 0 0 45.783 374.147 365.866 cm stream /F1 6 0 R 4.303 0.705 TD >> 2015 AP Chemistry free response 2a (part 1 of 2) 201 0 obj << /Meta133 147 0 R -0.012 Tc 0 G endobj /Meta58 72 0 R >> 0.003 Tw endobj 0000026777 00000 n >> 2.791 1.036 TD q /Font << /BBox [0 0 0.263 0.279] 135 0 obj << q >> stream stream /BBox [0 0 0.263 0.279] 45.289 0 0 45.355 81.303 493.844 cm Q /FormType 1 /Font << >> /Resources << q 0 -0.003 l stream ET q /BBox [0 0 9.507 1.562] endstream /BBox [0 0 9.507 1.795] 0 g Q 45.663 0 0 45.783 90.337 112.169 cm /F1 0.217 Tf /BBox [0 0 0.263 0.279] q ET /Type /XObject >> 0 g /Matrix [1 0 0 1 0 0] >> /F1 6 0 R 0.531 -0.003 l 0 w >> /Subtype /Form endobj 45.413 0 0 45.783 523.957 331.99 cm 0 0.279 m endobj /Meta136 150 0 R Correct 2.7 g Al is 0.10 mole, 100 mL of 1 M HCl is 0.10 mole. Q ET /Meta171 185 0 R 0 -0.003 l Q /Subtype /Form >> ET 578.159 332.743 l /Type /XObject /F1 0.217 Tf endstream Q >> Q 0.001 Tc Q 1.047 -0.003 l endobj /FormType 1 [( \(g\))] TJ /Resources << >> 9.775 -0.003 l q >> [(0)19(. 45.287 0 0 45.783 284.563 112.169 cm 0000017833 00000 n Stoichiometry example problem 1. /BBox [0 0 0.314 0.279] 0 0.279 m Q endobj 6. c In multiple choice questions without a calculator, you must look for the “easy math” − You will be most successful at this if ... 2Al ⎛ ⎝⎜ ⎞ ⎠⎟ 2g 1mol ⎛ ⎝⎜ ⎞ ⎠⎟ = 3 g H 2 7. c First you must realize this is a limiting reactant problem. 1.047 -0.003 l q /StemH 88 /Info 3 0 R /Meta174 Do >> /BBox [0 0 9.507 1.795] 0.015 w 1 g endstream Q /Matrix [1 0 0 1 0 0] q >> /Meta60 74 0 R /Type /XObject >> stream >> 0000034942 00000 n Practice Multiple Choice. /Length 122 /Font << Q q /Meta139 Do 0 0.279 m endobj 0 1.264 TD Q Q S /F1 0.217 Tf /FormType 1 0 1.562 m BT 1.3 mol of N 2. /Font << /Meta171 Do 0 g Q endstream 0.3803 mol = 37.1 g c) How many grams of the excess reactant will remain after the reaction is over? endobj /Resources << /Subtype /Form >> /Length 56 endobj /Length 571 >> 0 -0.003 l BT /Font << 0 G Q /Subtype /Form [( r)24(eac)22(t wi)37(th )15(7. BT 0000022984 00000 n /Meta110 124 0 R 0000027028 00000 n /Meta117 Do endstream q W* n 250 0 obj << W* n 0000015310 00000 n endobj 0 -0.003 l q 45.289 0 0 45.313 81.303 599.238 cm /FontFile2 23 0 R 232 0 obj << /Resources << /FormType 1 W* n BT 0.267 -0.003 l /Count 1 q endstream Q endstream /F1 6 0 R 0 1.795 m 0 -0.003 l [(C\))] TJ /Length 70 Q /F1 0.217 Tf 45.289 0 0 45.354 81.303 130.236 cm stream ET /Meta25 38 0 R /F1 0.217 Tf Q -0.002 Tc /Type /XObject Q Q [(2)] TJ Q 0.267 -0.003 l 0 w q /F3 0.217 Tf /F1 0.217 Tf 154 0 obj << /Matrix [1 0 0 1 0 0] 0.564 G 0.015 w 0 g /F1 0.217 Tf >> 0.564 G q Q 45.289 0 0 45.354 81.303 130.236 cm *Look back at preceding question. stream /Meta100 114 0 R 0000012582 00000 n q 0 -0.003 l /F1 6 0 R Q q /BBox [0 0 9.507 1.511] /Resources << /FormType 1 0 g /Meta197 211 0 R q /F1 6 0 R >> c.Calculate the mass of Fe 2 O 3 (s) produced. 1 J q q q q 86 0 obj << endobj /FormType 1 q q >> /Type /XObject endstream Q q /Type /XObject /Meta236 250 0 R >> 45.663 0 0 45.783 448.676 581.171 cm 0.005 Tc Q /F1 0.217 Tf endobj [( Al\(N)] TJ endobj /Meta102 Do >> 199 0 obj << /Resources << [(B\))] TJ Q 0 G 45.289 0 0 45.355 81.303 493.844 cm endobj /Type /XObject q endobj 0 0.279 m /Meta216 Do /StemH 88 /Meta3 Do Q /XObject << [(% y)-22(iel)-18(d )-18(of th)-31(e re)-20(ac)-17(tion )-18(is )-14(__)-15(__)-15(_)-16(___)-14(__)-15(.)] 0.564 G Q Q /Matrix [1 0 0 1 0 0] /Resources << /Meta76 90 0 R 0.267 0.279 l /Resources << /Type /XObject Q /Subtype /Form q 0 0.279 m 0.001 Tw 0 0.279 m Q stream q >> W* n Q >> /Length 56 >> q >> /Meta194 Do /Matrix [1 0 0 1 0 0] Q q 0.458 0 0 RG /Meta157 Do q Q 28 0 obj << q q >> 0.267 0.279 l BT /Type /XObject q 0 g endstream 259 0 obj << >> endstream >> /Matrix [1 0 0 1 0 0] endobj /Subtype /Form >> ____ 1. 0000003319 00000 n /Resources << /Meta126 140 0 R /Meta92 106 0 R q /Filter [/FlateDecode /DCTDecode] 0 G /Meta119 Do 191 0 obj << q 0000037793 00000 n /Meta139 153 0 R Q 252 0 obj << Q /Resources << Q /BBox [0 0 9.507 2.074] BT Q Q /Meta51 64 0 R /F1 0.217 Tf q [(26)] TJ ET endstream /Length 155 /FormType 1 0.267 0.279 l 0.031 0.083 TD 1 g /FormType 1 /Type /XObject 0 g W* n /Resources << >> /Meta40 Do endobj /Meta191 205 0 R /Meta241 255 0 R 45.289 0 0 45.355 81.303 493.844 cm >> ET Q >> 0000059966 00000 n W* n >> 0 -0.003 l /Subtype /Form /Meta11 19 0 R 3.275 0.418 TD /Meta29 42 0 R 0 g >> >> Q 0.001 Tw q stream /FormType 1 /Length 63 BT stream BT /Type /XObject endstream 0 0.279 m Q >> /FormType 1 /Subtype /Form 1 g 0000084128 00000 n /Meta218 232 0 R /Subtype /Form /Subtype /Form [(3)] TJ q W* n 0.267 -0.003 l q /Type /XObject /Meta22 Do 0000010560 00000 n 0 0.083 TD >> >> >> q >> endobj q /Font << q /Meta175 189 0 R a. 2.822 0.371 TD /Resources << 0.015 w stream 0 -0.003 l 0000065065 00000 n Q endobj /BBox [0 0 9.507 1.46] Q stream Q 4.22 1.367 TD stream >> Q 0 -0.003 l 0.267 0.279 l /F1 0.217 Tf BT /Matrix [1 0 0 1 0 0] endstream q 1 g 1.047 0.279 l Q >> /Font << 1.527 0.422 TD q >> q /Font << q 0 g 0.001 Tc /Resources << 141 0 obj << /Length 122 q q >> Q q ET /Length 122 q /Subtype /Form 0 g 0.267 0.279 l 0 w 0 g Q q stream Q 0000052810 00000 n 0 -0.003 l /Meta108 122 0 R >> /Type /XObject 0 g /Matrix [1 0 0 1 0 0] trailer Q BT ET q endobj 45.663 0 0 45.783 179.922 112.169 cm /F1 6 0 R 0000036821 00000 n Q Q BT -0.007 Tc /Type /XObject 45.289 0 0 45.355 81.303 493.844 cm endstream 1 g Q q 0 g /FormType 1 0.267 -0.003 l Q /Meta143 157 0 R 0 -0.003 l 0000028312 00000 n stream q /Meta76 Do 45.287 0 0 45.783 374.147 581.171 cm /Font << 248 0 obj << /F1 6 0 R /Meta161 175 0 R /F1 0.217 Tf 0.267 -0.003 l W* n [(O)] TJ Q endobj ET ET /Matrix [1 0 0 1 0 0] >> /Resources << Q >> /BBox [0 0 9.507 1.562] /Length 54 Q >> q 0.015 w 0000004064 00000 n >> W* n Q /FormType 1 >> endstream Q Learn vocabulary, terms, and more with flashcards, games, and other study tools. 0.002 Tc Determine the percent yield of the reaction when 77.0 g of CO 2 are formed from burning 2.00 moles of C 5 H 12 in 4.00 moles of O 2.. C 5 H 12 + 8 O 2 → 5 CO 2 + 6 H 2 O Q >> >> /Type /XObject 0.531 -0.003 l >> q /Resources << [(47)] TJ /FormType 1 0.267 0.279 l q 0000024431 00000 n 0 g /BBox [0 0 9.507 1.562] q >> Stoichiometry and Limiting Reactant Quiz Version A (PLEASE DO NOT WRITE ON THE QUIZ) Multiple Choice Identify the choice that best completes the statement or answers the question. 0 G [( \()20(s\))] TJ Q 33 0 obj << q Q >> endstream /XHeight 476 0000014809 00000 n Q q /Resources << /Subtype /Form 0 g endstream Q Q /Font << /Type /XObject Q 0 0.279 m /Meta130 144 0 R 0 g stream Q q startxref /Subtype /Form /BBox [0 0 9.507 1.46] q 4. /Font << /Length 76 /Meta173 187 0 R Q q 0 0.279 m /Type /XObject TJ endobj >> 0 g /Resources << /Resources << /Type /XObject 0 G /ItalicAngle 0 /FormType 1 >> 6.429 0.371 TD q 0 0.279 m Q Q 1 j /Type /XObject 1.047 0.279 l Q Q /Meta50 Do /Matrix [1 0 0 1 0 0] /FormType 1 0 g 45.287 0 0 45.783 36.134 682.048 cm 0 0.279 m q /F3 25 0 R Q >> BT /F1 6 0 R /Type /XObject /FormType 1 /BBox [0 0 0.263 0.279] /FormType 1 >> q q >> 1.047 0.279 l 0 G 0000042861 00000 n /BBox [0 0 9.507 1.562] /Type /XObject /Subtype /Form 158 0 obj << 0 0.083 TD /Type /XObject 45.663 0 0 45.783 448.676 365.866 cm Q /Meta166 Do endobj Q 0 g /Subtype /Form 0.004 Tw endstream endstream 0.267 -0.003 l /Matrix [1 0 0 1 0 0] /Subtype /Form Q 0.564 G 0 0.279 m endobj endstream /FormType 1 >> /Font << 0.314 0.279 l q 0 g 202 0 obj << BT /F1 0.217 Tf /Type /XObject >> q /Resources << 0000009844 00000 n [( \(g\))] TJ Q 0 G 0.267 -0.003 l /BBox [0 0 9.507 1.511] /Subtype /Form q /F1 6 0 R endstream >> /FormType 1 q endstream /Matrix [1 0 0 1 0 0] 1.047 -0.003 l ET • All questions are multiple choice. 2 0 obj << 2.527 0.752 TD [(Al)] TJ /Length 122 >> /Resources << S /FormType 1 /FormType 1 >> /Length 63 /Meta52 Do 0.015 w BT /Subtype /Form 56 0 obj << 45.289 0 0 45.354 81.303 130.236 cm stream >> endstream /Type /XObject /F1 0.217 Tf W* n 4.137 0.752 TD 0 g 0 0.279 m 45.663 0 0 45.783 448.676 475.777 cm >> /Length 68 /Length 62 BT Q ET -0.012 Tc Q Q 228 0 obj << 0.564 G q >> endobj /Type /XObject Q /BBox [0 0 1.047 0.279] /FormType 1 /BBox [0 0 11.968 0.279] 9.775 0.279 l 0 g 0000063236 00000 n /Subtype /Form q /BBox [0 0 9.507 2.074] 0 -0.003 l 0 -0.003 l /Font << 45.287 0 0 45.783 284.563 365.866 cm endobj /Subtype /Form Then solve the problem. 9.775 0 0 0.283 0 -0.003 cm 0 0.279 m /Meta200 Do -0.003 Tc 0.047 0.083 TD /F1 6 0 R /F1 0.217 Tf >> /Subtype /Form endobj endstream q endstream >> 1.877 1.036 TD /Subtype /Form 0 1.562 m q /Subtype /Form 0 g /Font << 0000067590 00000 n /Type /XObject /Font << ET 0000000011 00000 n /Subtype /Form [(O)] TJ /BBox [0 0 0.263 0.279] /Meta247 Do Q Q ET /BBox [0 0 1.047 0.279] Q /Length 62 /Type /XObject /Matrix [1 0 0 1 0 0] /FormType 1 1.047 0.279 l W* n 1 g q ET /Length 55 endobj /Type /XObject /Meta185 199 0 R q 0 G q /Contents [262 0 R] /Meta198 212 0 R /Meta38 51 0 R Q endobj endobj /Meta21 34 0 R /Matrix [1 0 0 1 0 0] 0000033895 00000 n /Subtype /Form /Font << >> /Type /XObject /F1 6 0 R q 9.775 -0.003 l /Resources << Q q /Type /XObject /F1 6 0 R -0.003 Tc BT q 0.346 0.083 TD W* n Q /Type /XObject /F1 0.217 Tf /Flags 32 BT 0 -0.003 l Q 0 -0.003 l q Q q 1.047 -0.003 l /BBox [0 0 9.507 1.795] q /BBox [0 0 9.507 1.46] >> q /FormType 1 /Matrix [1 0 0 1 0 0] /Font << Q >> /Meta118 Do /Matrix [1 0 0 1 0 0] /Type /XObject Q endstream ET 0.002 Tw >> /Meta6 14 0 R 0.047 0.083 TD 8.645 0.138 TD 0 0.279 m 1 J >> /Resources << q /Length 122 /Font << Q /Meta82 Do /F1 6 0 R 0.001 Tc 1.047 0.279 l Q /BBox [0 0 9.507 2.074] /Type /XObject q q endobj /Font << Q /Subtype /Form /Resources << /F1 0.217 Tf endobj >> Q [(p)-15(ol)21(lutan)16(t:)] TJ 0 w stream 3.338 0.753 TD BT /Meta37 50 0 R /Meta128 142 0 R /Resources << endstream 45.289 0 0 45.355 81.303 493.844 cm q q /Matrix [1 0 0 1 0 0] /Subtype /Form 0.458 0 0 RG 0 g >> /FormType 1 q /Matrix [1 0 0 1 0 0] 45.324 0 0 45.783 54.202 654.946 cm 3. /Subtype /Form The %)] TJ /Subtype /Form 129 0 obj << Q 0 g 0 g /BBox [0 0 0.531 0.279] Q 0000031721 00000 n W* n >> endstream endstream >> 145 0 obj << >> /Meta140 154 0 R [(O)] TJ /Meta127 Do q 1 g >> Q /F1 0.217 Tf 49 0 obj << 8.224 0.418 TD /Type /XObject endobj Q W* n /Type /XObject q stream Unit 3 Quiz--Limiting Reactants: Multiple Choice (Choose the best answer.) endobj W* n 0.031 0.083 TD ET q -0.001 Tw 1 g endstream Q Q 58 0 obj << >> Q 1.047 0.279 l BT Q 1.047 -0.003 l BT /Length 122 [(2)] TJ /Meta144 Do /Resources << 0 0.279 m q 0 G Q Q /Matrix [1 0 0 1 0 0] 0 G 24 0 obj << W* n [(2)] TJ /BBox [0 0 9.507 1.795] /Type /XObject ET stream 0.267 -0.003 l /Type /XObject Q /FormType 1 q Q /FormType 1 /Type /XObject /Resources << Q 0.001 Tc stream endobj /F1 6 0 R /Length 168 0 -0.003 l endstream /Type /XObject /Resources << 0.814 1.032 TD 0000003824 00000 n /Font << 31 0 obj << /Length 67 0000018340 00000 n 35 0 obj << /Resources << /F1 6 0 R q /DescendantFonts [<> /F1 6 0 R /BBox [0 0 1.047 0.279] Q Q /Font << /Type /XObject >> 59 0 obj << /Length 161 q /Meta23 Do q /Type /XObject 3.074 0.981 TD 45.289 0 0 45.355 81.303 493.844 cm [(3)] TJ The other reactants are partially consumed where the remaining amount is considered "in excess". /Type /XObject [(2\))] TJ 195 0 obj << 0 0.083 TD q [(3\))] TJ /BBox [0 0 9.507 1.511] -0.012 Tc 1.047 -0.003 l stream endobj Q ET /Meta86 Do endobj 193 0 obj << BT 1.047 -0.003 l /F1 6 0 R >> q /Meta33 46 0 R q BT /FormType 1 45.289 0 0 45.313 81.303 599.238 cm 45.413 0 0 45.783 523.957 654.946 cm 45.663 0 0 45.783 448.676 112.169 cm /Resources << /Font << The number of atoms in each compound in a chemical equation represent the a. masses, grams! 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