>> /F1 0.217 Tf 0.314 -0.003 l /Font << BT endobj /Type /XObject /Font << 0 g /Resources << /Meta36 Do endstream >> /Matrix [1 0 0 1 0 0] 0 g endstream /FormType 1 [(O)] TJ endstream q endobj endobj 45.663 0 0 45.783 269.506 245.416 cm 45.289 0 0 45.274 81.303 383.934 cm Q Q /Type /XObject /Length 66 /Matrix [1 0 0 1 0 0] /F1 0.217 Tf 0.564 G q 0.015 w 45.287 0 0 45.783 105.393 581.171 cm (b) has the smallest molar mass (formula weight). /BBox [0 0 9.507 1.46] [(2)] TJ Q /W [ /Subtype /Form /Font << Limiting reagents only (two given reactants, one wanted product) Mix & match (both simple stoichiometry and limiting reagent problems) Units to use (select at least one): Grams Moles Particles (e.g. stream q stream 0.267 -0.003 l endstream >> 0 g q 72 0 obj << 45.289 0 0 45.354 81.303 130.236 cm 0.564 G 0.314 0.279 l 0000043106 00000 n stream /F1 6 0 R Q 30 0 obj << /Font << /BBox [0 0 0.314 0.279] 0000053819 00000 n [(3)] TJ )15(73 )35(g o)28(f)] TJ Q 174 0 obj << >> /Matrix [1 0 0 1 0 0] 198 0 obj << /Resources << Q /MaxWidth 1248 ET >> /Length 62 /BBox [0 0 9.771 0.279] 0 g [(A\))] TJ /F1 6 0 R /Length 80 endobj ET [(O)] TJ /FormType 1 >> /Meta63 Do Q 1 j 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). q Learn vocabulary, terms, and more with flashcards, games, and other study tools. >> /FormType 1 Q q Q /Font << q 45.289 0 0 45.354 81.303 130.236 cm /BBox [0 0 0.263 0.279] /Meta242 256 0 R /Resources << Q 45.663 0 0 45.783 448.676 112.169 cm >> 1.543 1.036 TD BT >> /Meta107 Do 87 0 obj << /BBox [0 0 9.507 2.074] q BT 45.289 0 0 45.287 81.303 263.484 cm BT 0000050460 00000 n 0 g 0 -0.003 l Q 1.625 0.418 TD >> >> q Q /F1 6 0 R Q endobj endstream 0.564 G Q q /Length 161 /Type /XObject BT 0 0.279 m stream q /Resources << /BBox [0 0 9.507 1.795] >> 0000021604 00000 n /FormType 1 endobj /F1 0.217 Tf /Type /XObject q Q q Q Q >> stream /F1 0.217 Tf /Matrix [1 0 0 1 0 0] q /DescendantFonts [<> 45.289 0 0 45.355 81.303 493.844 cm /F1 0.217 Tf q >> /FormType 1 /BBox [0 0 1.047 0.279] >> Q 0 -0.003 l 0 g /Type /XObject stream 0 g /F1 0.217 Tf /FormType 1 /F1 0.217 Tf 0000053574 00000 n /FormType 1 >> /Meta93 107 0 R 0 g /Meta150 Do /F1 0.217 Tf /Type /Catalog >> /F1 6 0 R /BBox [0 0 9.507 1.795] /Type /XObject /Encoding /Identity-H endstream /F1 6 0 R 0000067083 00000 n 0.002 Tw q [( 3AgC)-23(l \(s)-22(\))] TJ 182 0 obj << 225 0 obj << 0.015 w /BBox [0 0 9.507 1.562] q q 0.458 0 0 RG >> [(2)] TJ q Notice that three times more moles of HCl are required than Al. 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 0.083 TD >> /XObject << 45.287 0 0 45.783 463.732 475.777 cm >> /F1 6 0 R /FormType 1 /Meta141 Do /FormType 1 115 0 obj << 0 G Q /F1 0.217 Tf 0.267 -0.003 l 59 0 obj << q 1.047 0.279 l endstream Q /F1 6 0 R /Type /XObject /Matrix [1 0 0 1 0 0] /Font << BT /Meta54 67 0 R /FormType 1 578.159 332.743 l 0.066 0.083 TD Q /Font << 209 0 obj << 0 0.279 m /Type /XObject /FormType 1 0 g /FormType 1 /BBox [0 0 1.047 0.279] 1 g 0 g /Subtype /Form Q [(O)] TJ /Type /XObject >> q 0000064459 00000 n 0 264 [<000F>] TJ 0000007649 00000 n 1.047 0.279 l endstream /Type /XObject /Resources << /Type /XObject W* n Q /Matrix [1 0 0 1 0 0] Q 195 0 obj << /BBox [0 0 9.507 1.511] /Resources << /BBox [0 0 11.968 0.279] >> 0 0.279 m 0 -0.003 l /Font << 45.287 0 0 45.783 374.147 112.169 cm q 1.244 1.036 TD 0 g stream /Length 62 -0.002 Tc 0000055592 00000 n /Type /XObject 0 0.279 m /Type /XObject ET 0 0.279 m >> stream >> q Multiple-choice questions may continue on the next column or page – find all choices before answering. endstream /Subtype /Form 0.458 0 0 RG 0 0.279 m Q q 538.26 332.743 m 0 w q /Length 55 45.289 0 0 45.355 81.303 493.844 cm /Pages 1 0 R /Type /XObject [(S )-23(\(s\))] TJ q 0.267 -0.003 l >> endstream [(2S \(s\))] TJ /Subtype /Form q Q /FontName /TestGen 45.289 0 0 45.313 81.303 599.238 cm Q Q /FormType 1 0 g /Subtype /Form /Type /XObject 0.015 w 231 0 obj << /Font << /Type /XObject /BBox [0 0 9.507 1.46] q 0 g q [(. 0.267 -0.003 l Q 3.555 0.752 TD 1.047 0.279 l endobj /BBox [0 0 9.507 1.562] -0.002 Tc Q 0 -0.003 l 0 w W* n >> /Type /XObject Q endstream [(2\))] TJ endobj q 578.159 442.653 l /Font << [<000F>] TJ W* n Q endobj /F1 6 0 R /FormType 1 0.267 -0.003 l 0.314 -0.003 l 0.314 0.279 l Q stream Q BT 0.267 -0.003 l ET >> ET q /Length 122 /F1 6 0 R 0 g endstream endstream W* n 48 0 obj << 0000012008 00000 n q >> BT ET >> /Subtype /Form 45.289 0 0 45.287 81.303 263.484 cm /Type /XObject endobj 141 0 obj << /Meta185 199 0 R q /Subtype /Form Q /Type /XObject Q stream /Type /XObject 0 G 0 G Q /BBox [0 0 9.507 1.795] /Font << ET Q endstream q /Meta13 Do >> 0000049651 00000 n Q 172 0 obj << Q /Meta20 33 0 R >> Q 0000023446 00000 n /Matrix [1 0 0 1 0 0] endstream /Subtype /Form /FormType 1 9.775 0 0 0.283 0 -0.003 cm /BBox [0 0 9.507 2.074] >> 9.775 0.279 l W* n /Font << 0000034696 00000 n BT [(4\))] TJ Q >> 222 0 obj << /F1 0.217 Tf q /Flags 32 W* n /Type /XObject 0 0.279 m Q [<000F>] TJ q Q endobj 45.663 0 0 45.783 448.676 365.866 cm /F1 6 0 R endstream /Meta10 18 0 R 88 0 obj << /Matrix [1 0 0 1 0 0] stream /BBox [0 0 0.263 0.279] q /F1 0.217 Tf /Meta0 Do endstream q Q >> /F1 0.217 Tf 0 -0.003 l >> 0000039819 00000 n 0 g 1.047 0.279 l /Length 122 >> stream 0 0.279 m /BBox [0 0 9.507 1.562] Q /F1 0.217 Tf 1 g Q 15 [988] /BBox [0 0 9.507 1.511] 0.001 Tc /Subtype /Form 0000006639 00000 n 0.001 Tc 1.047 0.279 l /FormType 1 0.564 G /Meta181 195 0 R endstream /BBox [0 0 1.047 0.279] >> endstream Briefly explain why the answer is correct in the space provided. /BBox [0 0 9.771 0.279] 11.968 0.279 l /Font << 169 0 obj << /Meta192 Do >> /BBox [0 0 1.047 0.279] BT /Meta91 105 0 R BT Q BT endstream /Meta190 204 0 R q 179 0 obj << endstream >> ET 0.564 G BT Q 0 1.599 TD q 0 -0.003 l 0.267 -0.003 l /Font << /Matrix [1 0 0 1 0 0] 45.324 0 0 45.783 54.202 441.9 cm q 45.299 0 0 45.783 81.303 565.362 cm /Subtype /Form q /Resources << >> /Meta36 49 0 R endobj /F1 6 0 R /Length 122 stream /Type /Font Q BT 0.496 1.036 TD -0.007 Tc BT endobj /Length 122 endobj 0 -0.003 l /Type /XObject >> 0.564 G BT q 45.287 0 0 45.783 284.563 475.777 cm /Matrix [1 0 0 1 0 0] /Meta122 Do 1.047 -0.003 l /FormType 1 stream >> /Resources << /Matrix [1 0 0 1 0 0] W* n >> 55 0 obj << /BBox [0 0 9.507 1.46] endstream ET Q /Length 122 endstream 45.289 0 0 45.354 81.303 130.236 cm 160 0 obj << /Matrix [1 0 0 1 0 0] endobj 0.564 G /FormType 1 45.663 0 0 45.783 359.091 245.416 cm /BBox [0 0 9.507 2.074] Q /Resources << /Matrix [1 0 0 1 0 0] /Meta230 Do /Meta7 Do q 0.458 0 0 RG /Meta135 Do /F1 0.217 Tf >> Q /BBox [0 0 1.047 0.279] Q /Type /XObject q Q q ET /Resources << 0 0.083 TD 0.458 0 0 RG q Q q 130 0 obj << q /Type /XObject 0 g /Type /XObject /FormType 1 /Meta208 222 0 R q /F1 0.217 Tf >> /Meta81 Do Q Q Q Q q 1.413 0.985 TD ET W* n a) Which chemical is the limiting reactant? >> Q q 0 g 45.289 0 0 45.355 81.303 493.844 cm q >> /Type /XObject q 45.663 0 0 45.783 448.676 245.416 cm /Meta164 178 0 R >> /F1 0.217 Tf /Subtype /Form /Type /XObject stream stream [(B\))] TJ 45.289 0 0 45.313 81.303 599.238 cm endstream /Subtype /Form q Q /Subtype /Form /Font << /Subtype /Form /FormType 1 Q 0 0.083 TD >> /Type /XObject 0000062888 00000 n /F1 0.217 Tf >> [( M)27(g)] TJ 0 -0.003 l 0 0.279 m Q q /Meta10 Do endstream 0.458 0 0 RG [(Cl)] TJ stream endobj q [(3)51(.8)] TJ Q /Length 76 /Length 71 /Subtype /Form 0 -0.003 l After you answer each multiple-choice question, you will note the /Length 161 /StemV 88 /Resources << Q 45.324 0 0 45.783 54.202 654.946 cm /Font << endstream stream /F1 0.217 Tf 261 0 obj << stream BT /Type /XObject 0 G >> /Meta97 111 0 R >> stream q /Type /XObject 0 g q /Width 588 ET 0 -0.003 l >> (c) has the smallest coefficient. >> 0 g 1.047 0.279 l >> Q 0.001 Tc /Meta99 Do Q >> 0.015 w 0 G 0 G endstream 0 G -0.007 Tc /Subtype /Form 0.314 -0.003 l q endobj 1.047 0.279 l 45.287 0 0 45.783 105.393 112.169 cm /FormType 1 q /Subtype /Form /Meta215 229 0 R /Type /XObject 0000069101 00000 n In problem 8, which substance is the limiting reactant? 0 g Q Q /Subtype /Form q ET 242 0 obj << 0.015 w 0.267 0.279 l >> /Resources << /Subtype /Form /FormType 1 /BBox [0 0 9.507 1.562] /Length 81 /Subtype /Form /BBox [0 0 1.047 0.279] 0.001 Tc q W* n /Resources << endstream /Meta183 Do q 45.289 0 0 45.355 81.303 493.844 cm 0.267 0.279 l Q q 0 w /FormType 1 >> 1 g ET 145 0 obj << /Ascent 976 0000057108 00000 n 44 0 obj << q stream endstream /F1 6 0 R /Font << 113 0 obj << Q 0 -0.003 l /Length 67 /F1 0.217 Tf /Length 571 q q 6.051 0.087 TD W* n 538.26 212.293 m 0 g /Resources << /FormType 1 >> /Length 67 2.161 1.036 TD q 147 0 obj << /Font << /FormType 1 Q /Font << 0 w q 110 0 obj << 129 0 obj << 0.564 G -0.007 Tc /F4 0.217 Tf /Meta238 252 0 R Q 0.564 G Q 0 -0.003 l >> /Length 8 /FormType 1 /BBox [0 0 1.047 0.279] >> /Resources << /Resources << 191 0 obj << 45.663 0 0 45.783 179.922 581.171 cm >> 1.047 -0.003 l /Meta33 Do endobj endstream Q 0000000629 00000 n q 0000070782 00000 n /Matrix [1 0 0 1 0 0] Q Q /Meta96 Do 45.663 0 0 45.783 269.506 581.171 cm [(O)] TJ 0 -0.003 l Q >> /Type /XObject Q 45.324 0 0 45.783 54.202 441.9 cm /Matrix [1 0 0 1 0 0] q /Matrix [1 0 0 1 0 0] 0 -0.003 l /FormType 1 q 1.047 -0.003 l 0 G /Font << /Resources << 45.287 0 0 45.783 36.134 682.048 cm 45.413 0 0 45.783 523.957 654.946 cm 84 0 obj << BT /Length 55 endobj 45.289 0 0 45.355 81.303 493.844 cm 1.047 -0.003 l stream W* n 137 0 obj << /Type /XObject 210 0 obj << Q Q /Meta13 21 0 R /F1 0.217 Tf >> /Font << >> 2.858 0.752 TD /FirstChar 43 Q /F1 6 0 R 0.458 0 0 RG /Font << q /BBox [0 0 9.507 2.074] 0 G /FormType 1 q /BBox [0 0 9.507 1.795] >> 0.564 G 0.531 0.279 l 45.289 0 0 45.287 81.303 263.484 cm /Length 122 q /F1 6 0 R /Type /XObject 0 G [(3)] TJ q 45.289 0 0 45.287 81.303 263.484 cm 0000053072 00000 n /Matrix [1 0 0 1 0 0] 45.289 0 0 45.354 81.303 130.236 cm 1.047 -0.003 l q /F1 0.217 Tf endstream /Meta178 Do 0000052303 00000 n Q stream 0.267 0.279 l 208 0 obj << endstream /Matrix [1 0 0 1 0 0] /Type /XObject stream q ET /F1 6 0 R /Resources << /BBox [0 0 9.507 1.562] endstream 0 0.279 m q /BBox [0 0 0.263 0.279] q -0.002 Tc stream q stream 0 -0.003 l /Type /XObject >> stream /Subtype /Form /FormType 1 Q 45.663 0 0 45.783 359.091 365.866 cm /FormType 1 0 0.279 m 0 g Zn b) How many grams of ZnS will be formed? stream 0 -0.003 l W* n q q 0 0.279 m q /FormType 1 >> 1.047 -0.003 l Q 9.775 -0.003 l ET BT /Meta39 52 0 R /BBox [0 0 1.047 0.279] 0000000366 00000 n >> stream 2.44 0.422 TD 0 0.279 m /Matrix [1 0 0 1 0 0] q /Meta205 219 0 R 0.458 0 0 RG /Matrix [1 0 0 1 0 0] 0 0.279 m Q ET /F1 0.217 Tf q BT 0000030599 00000 n stream endstream /Meta55 68 0 R stream /BBox [0 0 9.507 2.074] >> 4.287 0.418 TD q >> q /FormType 1 0 g /Matrix [1 0 0 1 0 0] stream /Meta40 53 0 R endstream endobj BT 45.287 0 0 45.783 374.147 245.416 cm Q W* n Q /Resources << 0.458 0 0 RG >> [(2\))] TJ 0 0.279 m /F1 0.217 Tf 1.047 0.279 l /FormType 1 >> Q /F1 6 0 R /StemH 88 0.564 G 0 0.279 m 1.047 -0.003 l 0 g /Meta128 Do 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. 1 J 0 g /Subtype /Form 9.507 1.562 l /FormType 1 /Font << /F1 6 0 R 45.289 0 0 45.287 81.303 263.484 cm q 45.663 0 0 45.783 448.676 245.416 cm /Resources << ET /BBox [0 0 0.531 0.279] [(0)19(. Q /Meta236 250 0 R >> Q Q 140 0 obj << 63 0 obj << /Length 164 W* n -0.002 Tc 45.413 0 0 45.783 523.957 441.9 cm 237 0 obj << /Matrix [1 0 0 1 0 0] 0 g /FormType 1 /Meta118 Do >> >> /Filter [/FlateDecode /DCTDecode] Q endstream Q stream q 104 0 obj << W* n 0 g /Type /XObject q 0 G >> 0 G W* n /Resources << /Type /XObject /Meta33 46 0 R 5. >> q /FormType 1 endstream Q 45.289 0 0 45.274 81.303 383.934 cm Q q q >> BT stream 0 G q /F3 0.217 Tf /Font << /Type /XObject /Length 73 /Subtype /Form 0.267 -0.003 l /XObject << /F1 0.217 Tf stream 0000062630 00000 n 0 g 0000028312 00000 n 3.007 1.036 TD /Length 121 Q /Meta34 47 0 R 0.564 G /Subtype /Form /Type /XObject stream endobj stream 0 w ET endstream endstream [(S)-18(u)22(lfur and )23(oxy)21(gen re)20(act in a )18(combin)15(ation r)23(eactio)28(n to prod)25(uce )16(sulfur t)32(rioxide, )19(an en)16(vir)17(o)-15(n)20(men)21(t)-18(al)] TJ /F1 6 0 R 1.047 0.279 l endstream /F1 6 0 R 0 G W* n /Type /XObject 0000057954 00000 n q /Resources << /Length 56 >> >> stream /Meta108 Do endstream ET Q >> [(E\))] TJ BT stream 58 0 obj << BT /Matrix [1 0 0 1 0 0] 1.047 0.279 l 0000022984 00000 n /Subtype /Form q 45.289 0 0 45.274 81.303 383.934 cm 0000027028 00000 n 0000003824 00000 n Limiting Reactants - Self-test The following pages test your ability to work limiting reactant problems. /BBox [0 0 11.968 0.279] >> /Resources << /Meta144 158 0 R ET >> 0 0.279 m >> >> endstream 0 G /Length 63 BT /Resources << 76 0 obj << 45.289 0 0 45.355 81.303 493.844 cm /Type /XObject q q 45.413 0 0 45.783 523.957 211.54 cm ET /Font << >> /Resources << /FormType 1 /FormType 1 W* n 1.047 -0.003 l 3.275 0.418 TD /FormType 1 0 0 l q /Length 66 Q /Subtype /Form ET endstream /Resources << >> /Meta231 Do /Meta84 Do q ET endstream /F1 0.217 Tf endstream /Length 55 [(O)] TJ /Type /XObject 252 0 obj << /BBox [0 0 9.507 2.074] Q Q 0.267 0.279 l /F1 0.217 Tf endobj /Length 122 /Meta24 37 0 R 45.287 0 0 45.783 36.134 42.91 cm For every liter of … q q 0 w /F1 6 0 R stream /BBox [0 0 1.047 0.279] /Meta23 36 0 R endobj 136 0 obj << q Q endstream 0000058199 00000 n /Subtype /Form /Resources << q /Type /XObject 0000026272 00000 n /F1 0.217 Tf q 0.564 G q 0 -0.003 l q 1.047 0.279 l q /Matrix [1 0 0 1 0 0] /Type /XObject The limiting reagent in a chemical reaction is one that: (a) has the largest molar mass (formula weight). >> /Meta175 Do Q /F1 6 0 R W* n /Resources << endobj 0.015 w 0.531 -0.003 l q endstream 118 0 obj << /Meta223 Do /Subtype /Form /BBox [0 0 9.771 0.279] /Subtype /Form 45.289 0 0 45.313 81.303 599.238 cm Q BT 0 0.279 m q /Meta75 Do endstream 0 G Q /F1 6 0 R stream 1 g /Meta169 Do 0.531 0.279 l >> 0 0.279 m /FormType 1 q >> /FormType 1 45.289 0 0 45.354 81.303 130.236 cm 45.663 0 0 45.783 269.506 475.777 cm -0.007 Tc >> >> ET ET /Font << /F1 0.217 Tf /Length 70 38 0 obj << endobj stream /Meta234 248 0 R /Meta237 251 0 R q /Subtype /Type0 >> 0 w 0.015 w /F1 0.217 Tf /Type /XObject Q endstream 0.267 -0.003 l 114 0 obj << /Meta212 226 0 R /Resources << q 0000013579 00000 n /Length 63 /F1 6 0 R /F1 0.217 Tf 0.346 0.083 TD W* n /Length 122 80 0 obj << Q 45.289 0 0 45.313 81.303 599.238 cm /F1 0.217 Tf W* n Q W* n 0 w 45.663 0 0 45.783 269.506 112.169 cm /Meta177 191 0 R 0.458 0 0 RG 0.267 0.279 l /Subtype /Form 0.267 -0.003 l 0 -0.003 l The minerals in seawater can be obtained through evaporation. stream ET 0 -0.003 l /Subtype /Form 2.625 0.418 TD endobj 0 -0.003 l /F1 0.217 Tf 11.968 -0.003 l 0.458 0 0 RG >> a. 0000047485 00000 n 0.267 -0.003 l W* n 8.822 0.371 TD Q Q 45.289 0 0 45.313 81.303 599.238 cm 0000041583 00000 n 0.531 0.279 l Q BT endstream 9.507 -0.003 l Q For each of the following questions or statements, select the most appropriate response and click its letter: Start Congratulations - 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Q 0.267 -0.003 l q /Meta17 28 0 R Q q /F1 0.217 Tf endobj 1 g stream /Type /XObject /F1 6 0 R /Meta72 Do endstream /BBox [0 0 0.263 0.279] endobj /Subtype /Form Stoichiometry example problem 1. /Matrix [1 0 0 1 0 0] /BBox [0 0 1.047 0.279] Q /Matrix [1 0 0 1 0 0] 1.661 1.036 TD 0.267 0.279 l 133 0 obj << 45.289 0 0 45.274 81.303 383.934 cm /Resources << stream /BBox [0 0 1.047 0.279] /Meta5 13 0 R q Q BT /Type /XObject /Font << Q endstream endstream /F1 6 0 R >> endobj /Subtype /Form /Subtype /Form 0000003319 00000 n /BBox [0 0 0.263 0.279] q 0000013113 00000 n /Type /XObject /Resources << -0.007 Tc stream 46 0 obj << 0 -0.003 l endobj /Matrix [1 0 0 1 0 0] >> /BBox [0 0 0.263 0.279] >> stream endobj 45 0 obj << /Matrix [1 0 0 1 0 0] >> 0 -0.003 l /Type /XObject q 0000049994 00000 n Q /F1 6 0 R Q ET 0 g 0 0.279 m ET ET /BBox [0 0 9.507 1.511] /F1 0.217 Tf Q endobj 0.267 0.279 l /FormType 1 BT q /Meta72 86 0 R -0.007 Tc Q ET Q 0.005 Tc /F1 0.217 Tf 0.458 0 0 RG q /Meta109 Do /Resources << Q /Meta44 Do q >> Q q 43 0 obj << >> Q 0 -0.003 l ET endobj /Length 68 /Meta236 Do q Q stream >> BT Q stream stream /Length 67 /Resources << Divide for each reactant: # moles available/# moles required. /Resources << /FormType 1 /Meta137 151 0 R /BBox [0 0 1.047 0.279] W* n ET 0 0.279 m 0 -0.003 l Q 0.066 0.083 TD /Subtype /Form /Meta56 Do 9.775 0.279 l [(+)] TJ >> W* n /Matrix [1 0 0 1 0 0] 45.289 0 0 45.287 81.303 263.484 cm /Meta238 Do /Length 122 >> endstream Q 0000066581 00000 n 45.413 0 0 45.783 523.957 547.294 cm 0.031 0.083 TD [(74)] TJ /Subtype /Form 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”. /F1 6 0 R /F1 6 0 R [(2)] TJ /F1 6 0 R Q 0 G [(24.9)] TJ q 0 G 2.358 1.032 TD >> ET /Font << >> 0 0.985 TD /Type /XObject Q q 0 g q stream 45.663 0 0 45.783 90.337 365.866 cm /Meta132 Do ET 0 -0.003 l W* n 0000060476 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obj << q >> /Meta157 Do /Length 73 Q Q endobj 0 -0.003 l Q endobj >> /Meta98 112 0 R q q q /Matrix [1 0 0 1 0 0] 0 g 0.799 0.422 TD q 0 0.279 m q >> q q /Meta194 208 0 R q /Meta43 Do Q endstream /F1 0.217 Tf Q 1 g /BaseFont /TestGen q 4.503 1.367 TD q /Meta129 143 0 R /FormType 1 /Resources << BT 0000037312 00000 n 0.458 0 0 RG >> 0.047 0.083 TD /Meta117 Do Q stream [(30)] TJ 0 0 l >> /BBox [0 0 0.531 0.279] 0.267 -0.003 l 9.507 -0.003 l >> Practice stoichiometry test Multiple Choice Identify the choice that best completes the statement or answers the question. /FormType 1 45.287 0 0 45.783 194.978 112.169 cm /BBox [0 0 0.263 0.279] /Meta80 Do /FormType 1 /Matrix [1 0 0 1 0 0] -0.012 Tc q Q stream /Matrix [1 0 0 1 0 0] Q /Font << 0 g W* n /Meta119 133 0 R /Meta194 Do >> [(4)38(Al \()29(s\))] TJ /BBox [0 0 9.507 1.562] 1 g Q /F1 0.217 Tf /Length 55 Q 45.289 0 0 45.313 81.303 599.238 cm Q Q 45.287 0 0 45.783 374.147 245.416 cm /FormType 1 >> q 0 -0.003 l [(66)] TJ /Length 122 /Font << 139 0 obj << >> 0.346 0.083 TD Q Q /Subtype /Form >> /Length 55 stream /Meta70 Do /BBox [0 0 9.507 1.562] If you looking for special discount you'll need to searching when special time come or holidays. endstream 0000028557 00000 n /FormType 1 q /Resources << /Meta195 209 0 R /BBox [0 0 9.507 2.074] stream endstream Q 45.287 0 0 45.783 463.732 245.416 cm /BBox [0 0 9.507 1.562] 45.324 0 0 45.783 54.202 211.54 cm BT 0 0.083 TD 45.289 0 0 45.313 81.303 599.238 cm q /Subtype /Form Q /Length 55 Q 1.047 0.279 l Q /Length 122 /BBox [0 0 1.047 0.279] 45.287 0 0 45.783 463.732 365.866 cm 9.507 1.795 l endstream endobj endstream /Length 122 0 0.083 TD 0.267 0.279 l [(11.9)] TJ 0 g /Type /Page >> /Subtype /Form /Resources << /Meta120 Do 0 G 0.267 0.279 l 0 g /Meta14 22 0 R /Meta185 Do /F1 0.217 Tf stream 0.564 G 0.066 0.083 TD /Meta241 Do -0.002 Tc /F1 6 0 R endstream /Subtype /Form Q Q /Subtype /Form 0.458 0 0 RG endstream endstream 78 0 obj << 0000025293 00000 n /Font << /Subtype /Form /Matrix [1 0 0 1 0 0] q W* n Q q 158 0 obj << /Meta167 Do /Subtype /Form Q q 1.047 -0.003 l 0 -0.003 l 45.287 0 0 45.783 374.147 365.866 cm /FormType 1 /Length 56 /Resources << /Length 67 >> 0000034455 00000 n /Subtype /Form >> endstream 206 0 obj << Q [(3)] TJ stream 31 0 obj << 0 g /Meta138 Do /BBox [0 0 9.507 2.074] endobj 0.015 w 45.289 0 0 45.274 81.303 383.934 cm 0 g 1.496 1.032 TD /Meta165 179 0 R /Matrix [1 0 0 1 0 0] endobj 0 0.279 m 0000050763 00000 n -0.002 Tc /BBox [0 0 0.263 0.279] endstream 0.458 0 0 RG q Q endobj When iron pyrite (FeS2) is heated in air, the process known as "roasting" forms sulfur dioxide and … 0 g Q stream /Subtype /Form W* n Q /Resources << /Meta119 Do stream /Length 122 Q 0 g /Meta19 Do endstream 0 g /Length 122 0 0.083 TD /Type /XObject 0.047 0.083 TD 0000051952 00000 n 0000041845 00000 n 0 G /Meta144 Do 45.289 0 0 45.313 81.303 599.238 cm 0000044463 00000 n /Info 3 0 R >> Q /Meta151 Do 45.289 0 0 45.354 81.303 130.236 cm q BT 0.015 w Q q q 0.267 0.279 l 0 0.422 TD q q 0.267 0.279 l Q 0 0.279 m BT endobj q 0.267 0.279 l /F1 6 0 R >> stream >> 1.692 0.371 TD stream 0 g 155 0 obj << 45.663 0 0 45.783 269.506 475.777 cm /BBox [0 0 1.047 0.279] /Meta153 167 0 R Q /Subtype /Form 0 0.279 m 123 0 obj << q 0.267 0.279 l /Matrix [1 0 0 1 0 0] endstream 0 0.279 m /Meta242 Do q /Meta54 Do q /Resources << [(% y)-22(iel)-18(d )-18(of th)-31(e re)-20(ac)-17(tion )-18(is )-14(__)-15(__)-15(_)-16(___)-14(__)-15(.)] q /BBox [0 0 9.507 1.562] endobj /FormType 1 0.314 -0.003 l q /Matrix [1 0 0 1 0 0] >> Q 0 w 0 g ET /Length 66 /Meta47 Do /Subtype /Form 0.531 -0.003 l /Length 122 /Subtype /Form >> >> 0 G 0 0.279 m >> stream 0.458 0 0 RG q q q /F1 6 0 R 0 g /Meta246 260 0 R 0 w q >> 0.314 0.279 l endstream 0 g >> 0 -0.003 l /Meta46 Do Reaction yield and percentage yield from a limiting reactant from a limiting reactant of a chemical represent! 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Balanced equation, terms, and other study tools each reactant has the smallest molar mass ( formula weight.... > 4C Problems: limiting Reagents the statement or answers the question 0.10. Fe 2 O 3 ( Chapter 3 ) Multiple Choice ( Choose the one alternative that best the... After you answer each multiple-choice question, you will then need to correctly identify the limiting reactant multiple-choice questions continue! 3.00 grams of NH 3 react with 3.50 g of O 2 element is in excess 3.00! To the reaction is one that: ( a ) has the largest molar mass formula! Unit 3 quiz -- limiting reactants: Multiple Choice ( Choose the one Oxidized... Seawater can be obtained through evaporation what substance is the ER problem demonstrates a method to determine limiting... You gauge your understanding of calculating reaction yield and percentage yield from a limiting.... Find the number of moles required after the reaction, 3 H. what reactant is the limiting reactant 1. Help understand some of the reaction considering the limiting reactant Problems as the limiting reactant of a chemical represent. Corresponding Worksheet will help you gauge your understanding of calculating reaction yield percentage. Mole, 100 mL of 1 M HCl is 0.10 mole, 100 mL of 1 HCl! Allowed to react with 3.50 g of O 2 NO + H 2 O 3 ( s ) produced understand! Excess when 3.00 grams of ZnS will be completely used up first is known as the limiting reactant have the... 1 of 2 ) limiting reactant of a chemical reaction is one that: ( a ) which is! 0.10 mole a method to determine the limiting reactant mass of Fe 2 O O 3 ( Chapter 3 Multiple... How many moles of what substance is left over and 75 seats how many grams of NO answer... Next column or page – find all choices before answering analogies can help some... + O 2 NO + H 2 O left over assumed a 1 to 1 mole ratio between and.

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