CHEM 1201 (Basic Chemistry) – Exam 2 Form 2 | Complete Prep Resource from LSU, Exams of Chemistry

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Exam 2 Form 2 MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Balance the chemical equation given below, and determine the number of grams of MgO are needed to produce 10.0 g of Fe203. MgO (s) + Fe (s)> Fe203 (s)+___ Mg (s) A)0.312g B)2.52g C)7.57g 3MQg0 + 2Fe > Fe,03 + 3Mg D) 0.841 g E) none of Mass of MgO these 1 mol Fe,0. 3molMgO 40.30gMgQ 10.0 g Fez,03 + Lien 9 gre =7.579 159.70 gFe,0,;\ 1molFe,0, _1molMgO 2) What is the empirical formula of a substance that contains 2.64 g of C, 0.887 g of H, and 3.52 g of O? Calculate moles of each element, then divide by the A) C2H402 smallest number of moles: Oxygen (O) and Carbon (C) Breese 2.64, 0.220 mol C C) CH4O Moles of C =>" = 0.220mol = 1 D) C3H404 mot , E) none of these __0887g 0.880 mol H _ Moles of H = 1.008 g/mol 0.880 mol 0.220 mol0 _ 352g 0.220 mol0 _ Moles of 0 = evn g/nal ™ 0.220 mol o7z0 mora. Empirical formula: CH,0 3) Lithium and nitrogen react in a combination reaction to produce lithium nitride: 6Li (s) + N3 (g) — 2Li3N (s) In a particular experiment, 1.50-g samples of each reagent are reacted. The theoretical yield of lithium nitride is g. A)7.5 Theoretical yield of LizN By ee 1mol Li) 2molLisN _ 34.83 9 LisN C)3.70 _(imol Li mol Lis. 83 g LisN _ 5 D) 2.51 150g ti(F oa gi) 6molLi * imolli,n 7519 1 E) 1.51 1.50 9 Nz ( 1 mol N, ) 2molLisN | 34.83 g Li;N = 3.73 g Li,N 28.02 g Nz. 1 mol Nz 1 mol LizsN Liis the limiting reagent 4) How many moles of BCI3 are needed to produce 10.0 g of HCl(aq) in the following reaction? BCl3 (g) + 3 H20 (I) — 3 HCI (aq) + B(OH)3 (aq) A) 0.823 mol B) 0.0914 mol 1mol HCl 1 mol BCL C) 0.274 mol 10.0 g HCl 3 — 0.0914 mol BCL D) 10.9 mol i 36.46 g HCL 3 mol HCl mores E) none of these 5) How many grams of CaCl2 are formed when 35.00 mL of 0.00237 M Ca(OH)2 reacts with excess Cl2 gas? 2 Ca(OH)2(aq) + 2 Cl2(g) — Ca(OCl)2(aq) + CaCl2(s) + 2 H2O0(1) A) 0.0217 g B) 0.00460 g C) 0.0184 g D) 0.00921 g E) none of these Mass of CaCl, 1LCa(OH), 0.00237 moles Ca(OH), 1molCaCl, 110.98 g CaCl, 35.00 mL Ca(OH) * To00mt CalOM)a TL Ca(OH); * Ymol Ca(OH), imolcaci, ~-°0460 9 CaCl, 6) What are the respective concentrations (M) of Cu2+ and CI- afforded by dissolving 0.666 mol CuCl in water and diluting to 522mL? A) 0.00128 and 1.28 B) 1.28 and 1.28 C) 0.784 and 0.174 D) 1.28 and 2.55 E) 0.00128 and 0.00128 Dissociation equation for CuCl,: CuCl,(s) > Cu?*(aq) + 2Cl- (aq) For every 1 mole of CuCl, that dissolves 1 mole of Cu* and 2 moles Cl are produced. moles of solute Molarity (M) = volume of solution (L) 0.666 mol CuCl, 1mol Cut 24) = (eu*) 0.522L 1molCuCl, = 1.28 M Cu" 0.666 mol CuCl, 2 mol Cl- [CU l=——Qea9 * imol CuCl, =2.55M Cl" 10) How many milliliters of 0.550 M hydriodic acid are needed to react with 15.00 mL of 0.217 M CsOH? HI (aq) + CsOH (aq) — CsI (aq) + H20 (I) A) 38.0 mL B) 5.92 mL C) 0. 0263 mL D) 0.169 mL E) none of these 1LCsOH .217 mol CsOH 1 mol HI 1LHI 1000 mL HI 15.00 mL CsOH = 7y00 mi CsOH” 1LCsOH * imolCsOH” 550molHIiLHi "22 ™/ #7 11) What volume (mL) of a 4.39 M lead nitrate solution must be diluted to 849.7 mL to make a 1.16 M solution of lead nitrate? A) 0.00599 M2V, B) 0.00445 Y= 4 C) 3220 1 D) 4330 (1.16 M)(849.7 mL) E) 225 Vi = —jn0y 1 4.39M V, =224.5mL ~ 225 mL 12) How many of the following compounds are soluble in water? Cu(OH)2 KNO3 NH4Cl — Li2S A)3 Cu(OH)2: Insoluble B)4 KNO: c)2 '3: Soluble D)O NH4CI: Soluble E)1 LigS: Soluble 13) If 100. mL of 0.800 M Na2SO4 is added to 200. mL of 1.20 M NaCl, what is the concentration of Na+ ions in the final solution? Assume that the volumes are additive. E are i Moles of Na2S0O,4 = 0.800 M x 0.100 L = 0.0800 mol C) 1.07M Moles of [Na*] from Na2SO4 = 2(0.0800mol) = 0.160 mol D) 1.33 M Moles of NaCl = 1.2 M x 0.200 L = 0.240 mol E) none of these Moles of [Na‘] from NaCl = 0.24 mol Total moles of [Na*| = 0.160 mol + 0.240 mol = 0.400 mol Total volume of solution: 0.300 L Concentration of [Na*| = 0:400'mol ~1.33M “t= 0300L ~ 14) How many molecules of HCI are formed when 80.0 g of water reacts according to the following balanced reaction? Assume excess ICI3. 2 IClg +3 H2O > ICI + HIO3 + 5 HCI A) 5.36 x 1024 molecules HC! B) 2.68 x 1024 molecules HCI C) 4.46 « 1024 molecules HCI D) 4.46 x 1025 molecules HCI E) 8.04 x 1024 molecules HCI 23 80.0 g H20 1molH,0_ , SmolHCl , 6.022 x10? molecules = 24, 18.019H20 3molH20 1 mol HCL = 4.46 x 10° molecules HCL 15) Methane (CH,) and oxygen react to form carbon dioxide and water. What mass of water is formed if 9.6 g of methane reacts with 38.4 g of oxygen to produce 26.4 g of carbon dioxide? A) 26.4g B) 48.0g C) 44.49 D) 21.6g E) none of these CH, +20, > CO, + 2H,0 9.69 CH, ( 1mol CH, ) 2 mol H20 | 18.02 g H20 16.04gCH,)* 1molCH,” 1molH,o ~ 24:69 #20 38.49 0+ (a2) 2molH,0 18,02 9H, 0 w090,)" = 21.6 g H,0 2molQ, 1molH,0 Neither is a limiting reagent 16) How many Cu atoms are contained in 896 g of Cu? A) 8.49 x 1024 Cu atoms B) 4.27 x 1022 Cu atoms C) 5.90 x 1025 Cu atoms D) 4.18 x 1024 Cu atoms E) 7.09 x 1021 Cu atoms 1molCu 6.022 x 107 Cu atoms * 63.546 g Cu 1mol Cu 896 9 Cux = 8.49 x 1074 Cu atoms