/Font << /Meta49 Do 170 0 obj << Q >> /Meta246 260 0 R 0.066 0.083 TD q Q q /Meta135 149 0 R q 0 -0.003 l /FormType 1 endstream W* n Q /Length 122 45.289 0 0 45.313 81.303 599.238 cm 0 g 245 0 obj << Q /Length 67 [(3)] TJ /Subtype /Form 43 0 obj << endobj q 45.287 0 0 45.783 194.978 245.416 cm /Resources << endstream /Type /XObject stream /Type /XObject Q 0 0.279 m /Subtype /Form 0.047 0.083 TD /Meta85 Do Practice stoichiometry test Multiple Choice Identify the choice that best completes the statement or answers the question. >> >> 0000014809 00000 n 0000047008 00000 n /Subtype /Form 0 0.279 m /Subtype /Form endobj 0000033360 00000 n q q 0 0.279 m 45.289 0 0 45.274 81.303 383.934 cm >> Chapters 1-3 and 22: 20 Multiple-choice questions; 4 Free Response Questions (EF, dimensional analysis, stoichiometry, limiting reagents, theoretical yields, percent yield, reactions, periodic table and nuclear formulas). 45.663 0 0 45.783 269.506 365.866 cm 0.015 w Q /F1 0.217 Tf stream /FormType 1 0 w 0000037549 00000 n -0.003 Tc Q endobj /Meta141 Do /Matrix [1 0 0 1 0 0] endobj q /Type /XObject /Meta231 245 0 R Q /F1 0.217 Tf 0.001 Tc >> Q Q 45.289 0 0 45.287 81.303 263.484 cm 0 0.279 m 45.289 0 0 45.355 81.303 493.844 cm /Length 168 q q /Font << /F1 0.217 Tf Q >> 0000002668 00000 n [( \(g\))] TJ q 9.775 0.279 l Q endstream xÚûãÿ/7O7FFF dø›ÁÈ£ü?ÀÀÍÁ ÈäÃÈ(Èğÿƒ021³°²±sprql``bdffbafeeaÊÖåXY… لٕ [( \(aq\))] TJ /Meta132 Do stream 4.22 1.367 TD ET >> >> q /BBox [0 0 9.507 2.074] q q nanni (arn437) – Limiting Reactants and Percent Yields - video – CNS – (87654) 1 This print-out should have 2 questions. Q /F1 0.217 Tf stream -0.001 Tw 1.047 0.279 l W* n Q /Subtype /Form /BBox [0 0 9.507 1.511] q stream q /F1 0.217 Tf 0 w ET 0 g /BBox [0 0 9.507 1.795] 45.287 0 0 45.783 374.147 365.866 cm W* n 0 G /Meta9 Do q In problem 8, which substance is the limiting reactant? 0000067083 00000 n endstream stream 0.015 w endobj /BBox [0 0 9.507 1.795] 194 0 obj << 0 g /Type /XObject 0000055837 00000 n endobj stream BT /BBox [0 0 0.263 0.279] /Meta12 20 0 R >> 0 G 0 -0.003 l 0 G >> 45.289 0 0 45.313 81.303 599.238 cm q /Meta226 Do 0.267 0.279 l q q Q 0 0.279 m /Type /XObject The %)] TJ 0000057348 00000 n Q q 0 w 0000048923 00000 n W* n /BBox [0 0 1.047 0.279] /BBox [0 0 1.047 0.279] 0.564 G /Meta103 Do 0.267 0.279 l /Meta156 170 0 R 0000061187 00000 n 0 -0.003 l /BBox [0 0 0.263 0.279] BT /Meta119 Do 0000066581 00000 n 214 0 obj << Q /Matrix [1 0 0 1 0 0] /Meta69 Do W* n 0 w /BBox [0 0 1.047 0.279] /Resources << 45.289 0 0 45.287 81.303 263.484 cm stream This is the currently selected item. Q 0 0.279 m stream >> 203 0 obj << [(O)] TJ 0.267 -0.003 l q /Meta6 14 0 R 0.015 w /Type /XObject >> ET 0 G W* n 0.267 0.279 l >> /Resources << Q ET Q 101 0 obj << 0 w 1.047 0.279 l >> stream 109 0 obj << q 1 g /F1 0.217 Tf /BBox [0 0 9.507 1.795] /Meta109 Do endobj /F1 6 0 R /Length 62 stream /F1 6 0 R 0 g 0.015 w >> q /F1 6 0 R ET /Type /XObject /Resources << Q [(\()-25(O)-25(H\))] TJ /Length 63 Q /F1 0.217 Tf /Type /XObject q Q Q /F1 0.217 Tf 0 G >> 36 0 obj << /Resources << >> 0 -0.003 l endstream q 0 G [(2)] TJ 0 w 1 g Q /Subtype /Form /Subtype /Form 8.311 0.422 TD 0000006377 00000 n BT /Matrix [1 0 0 1 0 0] /FormType 1 0.001 Tc Q /Length 122 45.289 0 0 45.313 81.303 599.238 cm q /FormType 1 0000030599 00000 n 45.287 0 0 45.783 374.147 475.777 cm 9.775 -0.003 l endstream /Widths [ 500]>> 0.015 w 223 0 obj << 0000038259 00000 n q -0.002 Tc 1.047 -0.003 l endobj >> >> /Resources << 0 G q Which substance is the limiting reactant when 8.0 g of sulfur reacts with 12 g of oxygen and 16 g of sodium hydroxide according to the following chemical equation: 2 S(s)+ 3 O 2 (g)+ 4 NaOH(aq)→ 2 Na 2 SO 4 (aq)+ 2 H 2 O(l) A) S(s) B) O 2 (g) C) NaOH(aq) D) None of these substances is the limiting reactant. /Matrix [1 0 0 1 0 0] stream /Resources << 0 0.279 m /FormType 1 q /Matrix [1 0 0 1 0 0] q /Matrix [1 0 0 1 0 0] /FormType 1 0.564 G q Q 0.015 w 1.047 0.279 l BT 0.458 0 0 RG 0 -0.003 l /F1 0.217 Tf ET /BBox [0 0 0.263 0.279] S 45.324 0 0 45.783 54.202 441.9 cm /FormType 1 0 -0.003 l /Matrix [1 0 0 1 0 0] Q Q 0 264 q stream Thus, HCl is the limiting reagent. /BBox [0 0 9.771 0.279] /FormType 1 /Subtype /Form 2.574 1.032 TD BT q >> stream 0 g /Font << /Matrix [1 0 0 1 0 0] Q 0 -0.003 l /FormType 1 /Resources << /Matrix [1 0 0 1 0 0] 166 0 obj << endstream /Meta160 Do /Font << /Subtype /Form Q 42 0 obj << endstream W* n >> 0.015 w /Type /XObject /FormType 1 endstream /Meta97 Do q /F3 25 0 R /Meta198 Do endobj 0000062164 00000 n -0.001 Tc Q q >> endobj /BBox [0 0 9.507 1.795] 1.047 -0.003 l endstream /Meta206 Do /FormType 1 BT /Matrix [1 0 0 1 0 0] 0 g /Type /XObject /BBox [0 0 1.047 0.279] Q /BBox [0 0 0.263 0.279] q 0000051230 00000 n endstream /F1 6 0 R /Matrix [1 0 0 1 0 0] 45.289 0 0 45.354 81.303 130.236 cm /Meta217 231 0 R /Length 62 q stream /Meta52 65 0 R [(21)] TJ /F1 6 0 R /Meta231 Do /BBox [0 0 9.507 1.511] BT BT q >> /Length 71 /Font << q Q /Meta81 95 0 R /Matrix [1 0 0 1 0 0] /Meta193 207 0 R -0.007 Tc >> q /Font << endstream 45.413 0 0 45.783 523.957 441.9 cm /FormType 1 endstream When the formula equation is correctly balanced the coefficient of Fe is number a. 0000024431 00000 n /Font << endobj ET >> /Meta239 Do 0 g Q /Meta243 257 0 R 0 g /Subtype /Form stream /Parent 1 0 R 1 g endstream 45.287 0 0 45.783 463.732 112.169 cm Q Incorrect You probably assumed a 1 to 1 mole ratio between reactants and products. /BBox [0 0 9.771 0.279] >> stream endobj 0000022244 00000 n 0 g /Type /XObject 2.192 0.752 TD q 0 g 0.001 Tc /Resources << /Type /XObject Empirical formula from mass composition edited. Q q /Meta234 248 0 R /F1 6 0 R >> /Matrix [1 0 0 1 0 0] 0000000121 00000 n q 45.663 0 0 45.783 90.337 365.866 cm 1.047 -0.003 l 0 G >> /Matrix [1 0 0 1 0 0] endobj endobj /FormType 1 >> Q >> stream /Resources << /Meta62 Do endstream 1.047 -0.003 l 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). /Subtype /Form /Meta100 114 0 R /F1 6 0 R 0 G 185 0 obj << /Matrix [1 0 0 1 0 0] Q 0000027796 00000 n >> /Matrix [1 0 0 1 0 0] 0.001 Tc 0 0.083 TD 0.531 0.279 l >> 45.289 0 0 45.354 81.303 130.236 cm 78 0 obj << Q BT 0.458 0 0 RG 0 -0.003 l /Meta68 Do endstream 128 0 obj << /Meta140 Do /F1 6 0 R /StemV 88 The mole and Avogadro's number. stream >> Q /F1 0.217 Tf 0 0.083 TD 0 g /Matrix [1 0 0 1 0 0] endstream stream /Subtype /Form /Length 122 Q /BBox [0 0 9.507 1.562] BT BT /F1 0.217 Tf Q 0 0.083 TD /Length 187 45.287 0 0 45.783 194.978 581.171 cm endstream 0 0.083 TD Q q 0 g 156 0 obj << /Length 155 Want to master theoretical yield? /Type /XObject endobj BT [(2\))] TJ endstream /F1 6 0 R >> >> /Resources << q /Meta145 Do (e) is in excess. 84 0 obj << /Meta232 246 0 R /Matrix [1 0 0 1 0 0] 0 -0.003 l /Matrix [1 0 0 1 0 0] 0 0 l EŒ'.äPvÚxP4èâ㤢&N.1q I)U5u M-S3sK+gW7wO¯à�Ğ°ğˆÈ¨ä”Ô´ôŒÌ¬â’Ò²òŠÊªæ–Ö¶ö�ήI“§L�6}ÆÌY‹/YºlùŠ•«6m޲uÛö;w:|äè±ã'N�ºtùÊÕk×oܼõğÑã'OŸ=ñòÕÇOŸ¿|ıöıÇÏ_ÿo1p€ücÏğïŸÆ¢®!€Öÿ¿ Áh®endstream 0 w /BBox [0 0 0.314 0.279] Q Q /FormType 1 >> 0.458 0 0 RG 240 0 obj << >> Q >> 0 0.279 m q 0 0.279 m /Matrix [1 0 0 1 0 0] 0 2.074 m 0 G 0.314 -0.003 l ET Practice Multiple Choice. >> /Meta214 Do 45.287 0 0 45.783 463.732 245.416 cm 0000068354 00000 n /F1 0.217 Tf 0000037054 00000 n endstream 0 g q /F1 0.217 Tf /Subtype /Form 0.564 G Q /Meta211 Do 0 G BT 0.458 0 0 RG 104 0 obj << stream 0000012814 00000 n 0 g BT /Meta229 Do 0 -0.003 l 79 0 obj << 8.645 0.138 TD 0000065327 00000 n [( via th)15(e e)15(quation )16(be)16(low?)] -0.002 Tc >> /BBox [0 0 0.263 0.279] >> Q /Length 55 stream /BBox [0 0 9.507 2.074] q Limiting reactant and reaction yields. /BBox [0 0 9.507 1.46] 0 0.087 TD 0.267 -0.003 l Q 0 g q stream q 1.047 -0.003 l 2.94 0.422 TD /Font << Q 45.663 0 0 45.783 359.091 365.866 cm 0 G Q q Q >> /FormType 1 >> 45.287 0 0 45.783 36.134 682.048 cm 0.458 0 0 RG 0.564 G 45.287 0 0 45.783 194.978 581.171 cm /Meta2 Do 0.564 G /F1 6 0 R endstream /F4 32 0 R Q q ET /Length 55 /Matrix [1 0 0 1 0 0] /F1 6 0 R Q endstream >> /Matrix [1 0 0 1 0 0] Question: Multiple Choice: In The Reaction: C + 2Cl2 --> CCl4, Chlorine Is The A. Oxidizing Agent B. -0.007 Tc /Meta41 54 0 R 45.289 0 0 45.287 81.303 263.484 cm /Meta201 215 0 R /Type /XObject >> /Matrix [1 0 0 1 0 0] q q /FormType 1 BT 0 G -0.002 Tc /Font << /Resources << 1.047 -0.003 l q /Type /XObject /Resources << 1.047 -0.003 l endobj 0.346 0.083 TD /Length 122 ET 40 0 obj << 1.047 -0.003 l /Resources << 0.015 w ET 0000065065 00000 n 0.015 w /CapHeight 476 1.377 1.032 TD /Subtype /Form 192 0 obj << endobj q /Meta172 186 0 R /Meta38 51 0 R /Matrix [1 0 0 1 0 0] endobj /Meta195 Do TJ 99 0 obj << /BBox [0 0 0.263 0.279] 0000037793 00000 n /Meta26 39 0 R 0 G 0 -0.003 l endobj /F3 0.217 Tf /BBox [0 0 1.047 0.279] /F1 0.217 Tf endobj endobj W* n [(O)] TJ /Length 121 0 -0.003 l stream 45.287 0 0 45.783 374.147 112.169 cm /Length 122 /Subtype /Form BT 9.507 0 l /BBox [0 0 9.507 1.46] stream [(D\))] TJ 0.047 0.083 TD endobj 45.324 0 0 45.783 54.202 547.294 cm 45.289 0 0 45.355 81.303 493.844 cm endobj /Matrix [1 0 0 1 0 0] Q 0 -0.003 l endstream ET /Meta191 205 0 R 173 0 obj << 58 0 obj << 0 G 0000051696 00000 n 0 0.279 m /Meta180 Do 0.267 0.279 l 45.289 0 0 45.287 81.303 263.484 cm >> /F1 0.217 Tf >> Briefly explain why the answer is correct in the space provided. /Subtype /Form W* n /Meta163 177 0 R /Subtype /Form /Meta17 Do q /Subtype /Form BT 0 g ET /Font << endstream /FormType 1 [(H)] TJ /F1 0.217 Tf Q /FormType 1 endstream stream /Length 122 /F1 6 0 R /BBox [0 0 1.047 0.279] /Meta221 235 0 R 200 0 obj << /Matrix [1 0 0 1 0 0] 0 -0.003 l -0.005 Tc 0000055090 00000 n 81 0 obj << /Resources << /FormType 1 Q 0 g W* n 1.047 0.279 l Q endobj /FormType 1 1.047 -0.003 l 0.267 -0.003 l BT >> stream 45.289 0 0 45.274 81.303 383.934 cm 0.015 w /Meta130 144 0 R 1.047 -0.003 l /Type /XObject q /Type /XObject 45.289 0 0 45.287 81.303 263.484 cm 0.001 Tc Q /Type /XObject Q Q 1 j /F1 0.217 Tf /Meta130 Do 0.458 0 0 RG /Subtype /Form 45.324 0 0 45.783 54.202 211.54 cm 0000035662 00000 n q /F1 0.217 Tf q endobj 0000019889 00000 n 0 -0.003 l Q >> q Q >> /Resources << q Q /Font << /Resources << >> /Matrix [1 0 0 1 0 0] /Meta137 Do BT 1 j >> 45.289 0 0 45.354 81.303 130.236 cm endstream -0.003 Tc >> /Resources << q 45.287 0 0 45.783 374.147 245.416 cm q Q 96 0 obj << 0 g q endobj q >> /Type /Pages q Q Q q /Matrix [1 0 0 1 0 0] [(A\))] TJ /Resources << 6.429 0.371 TD Q endstream /FormType 1 174 0 obj << /Meta140 154 0 R endobj q /Meta60 Do [( 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 /Subtype /Form TJ 45.289 0 0 45.354 81.303 130.236 cm 0 G 0.015 w The minerals in seawater can be obtained through evaporation. Worked example: Calculating the amount of product formed from a limiting reactant. /Font << /Matrix [1 0 0 1 0 0] 0 -0.003 l Q q q >> 3.24 0.989 TD /Resources << stream 0000015554 00000 n 56 0 obj << trailer 0 g 0 0.279 m >> 0.458 0 0 RG 0000001988 00000 n Q 1 g /FormType 1 Q /Meta47 Do [(C\))] TJ /FormType 1 Q /BBox [0 0 0.263 0.279] 0000042599 00000 n BT /Type /XObject q /Resources << Q [(In a p)34(articu)19(lar ex)18(perim)18(ent, t)18(he r)19(eaction o)27(f 2.5 g )22(o)-15(f A)18(l with 2)22(.5 g of)] TJ >> /FormType 1 /Type /XObject /F1 6 0 R q q 0 -0.003 l )18(0 )-16(g)] TJ Q W* n /F1 6 0 R 169 0 obj << /Length 71 0000011737 00000 n Q /F1 0.217 Tf 172 0 obj << /Matrix [1 0 0 1 0 0] /Meta125 139 0 R q W* n Q 45.289 0 0 45.287 81.303 263.484 cm Q 0 -0.003 l /F1 6 0 R Q /Font << -0.001 Tc 0000018079 00000 n W* n 0000043612 00000 n 45.289 0 0 45.313 81.303 599.238 cm /Meta192 Do Q /Meta8 Do >> 0 g 0 0.279 m >> W* n q endobj Q endobj >> /Font << /Type /XObject 0 g q >> q 45.289 0 0 45.274 81.303 383.934 cm 0 g Q /FirstChar 43 ET stream Q /Subtype /Form q /Length 122 /Meta68 82 0 R stream q /Meta233 247 0 R Q 0.346 0.083 TD >> q 45.289 0 0 45.313 81.303 599.238 cm The)] TJ 0 g 0000038026 00000 n /Subtype /Form Q /Length 55 0.267 -0.003 l q 45.663 0 0 45.783 179.922 245.416 cm 0000060710 00000 n /BBox [0 0 9.507 1.511] 0000038544 00000 n /Matrix [1 0 0 1 0 0] endobj Q endobj 45.287 0 0 45.783 105.393 475.777 cm /Subtype /Form 0 g 0000049156 00000 n 45.289 0 0 45.354 81.303 130.236 cm ET ET /F1 6 0 R Q /Length 55 W* n endstream ET endobj 0000070110 00000 n /Subtype /Form q /Meta137 151 0 R /Meta20 33 0 R /Subtype /Form /F1 6 0 R 9.507 1.562 l /Meta106 120 0 R /FormType 1 Q /Meta205 219 0 R /Length 55 Q /Meta99 113 0 R /Meta235 Do endstream 1.047 -0.003 l 0000019085 00000 n [(O)] TJ 0000025293 00000 n Q 0000020655 00000 n stream q Q /F1 0.217 Tf Q q q -0.001 Tw endobj 578.159 548.047 l q 45.287 0 0 45.452 36.134 673.014 cm >> /Resources << /Meta192 206 0 R /Type /XObject >> >> q 9.775 0.279 l >> /F1 0.217 Tf /F4 0.217 Tf >> 0 G Q Q /Length 122 endobj /Resources << /F1 6 0 R 1 g W* n 45.289 0 0 45.355 81.303 493.844 cm 0 G /Resources << 0.267 0.279 l /Type /XObject Q /Length 122 /FontDescriptor 24 0 R Q >> q q q 45.287 0 0 45.783 463.732 365.866 cm 8.661 0.422 TD BT q /Length 66 q 0 -0.003 l 9.775 -0.003 l /Meta217 Do /F1 6 0 R ____ 1. Q endstream q 1.047 0.279 l q /F1 6 0 R Q BT 0 -0.003 l Q /FormType 1 144 0 obj << /Meta75 Do /FormType 1 /Resources << Q q 0.531 0.279 l Q 45.289 0 0 45.354 81.303 130.236 cm Q /Subtype /Form /F1 0.217 Tf endstream /Matrix [1 0 0 1 0 0] /Font << 0.001 Tc stream stream 0 w q Q /Meta202 216 0 R /Type /XObject endobj 8.94 0.422 TD W* n /Meta110 124 0 R /Subtype /Form >> 1.047 -0.003 l W* n endobj q endstream /Ascent 976 /FormType 1 /F1 0.217 Tf BT ET /FormType 1 q W* n /Length 122 /BBox [0 0 9.507 1.562] 0000015310 00000 n /Font << Q 0 0.279 m >> (c) The following equation for the reaction involving hypothetical substances, A, B, C, and D, implies that the products C and D are always produced in a three to one mole ratio. >> 0 G 0 g >> 45.289 0 0 45.355 81.303 493.844 cm 0 g BT endobj /Meta4 12 0 R /Font << /Meta35 48 0 R 0 0.418 TD q 0.531 -0.003 l 0 -0.003 l 0000084128 00000 n 0.267 -0.003 l /F1 0.217 Tf Q /F1 6 0 R /Type /XObject >> /Meta160 174 0 R /Length 161 endstream Q /BBox [0 0 9.507 1.46] Q /Resources << /Meta42 55 0 R /Font << 0 G >> /Meta138 152 0 R /Meta99 Do [( S)] TJ W* n /Type /XObject >> q 1.047 -0.003 l endobj >> endstream 0.267 -0.003 l 142 0 obj << >> /F4 0.217 Tf /Meta126 140 0 R /Length 55 ET /Type /XObject /F1 0.217 Tf /FormType 1 0 -0.003 l q /Subtype /Form /Length 223 /Meta181 Do 0.001 Tc 1.047 0.279 l >> >> One alternative that best completes the statement or answers the question 8, which substance is over... 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