in a Molar Mass/gmor CHOH 46.07 50. A particular yeast ferments 180.2 glucose (C

Question

in a Molar Mass/gmor CHOH 46.07 50. A particular yeast ferments 180.2 glucose (C HO) and generates CHiO ethanol (C H OH) with an 87.0% yield. What mass of glucose is require 475 g of ethanol by this process? (B) 929 g (A) 808 g (C) 1.07x10'g (D) 2.14x10'g A reaction mixture initially contains 0.250 mol of Zns and 0.250 mol of O2. After the mixture has reacted completely, how many moles of the excess reactant are 5 I. left? 22nS(s) + 301g) 2ZnO(s) + 2so,(g) (B) 0.167 mol ZnS (D) 0.250 mol O (A) 0.083 mol ZnS (C) 0.125 mol O what is the net ionic equation for the reaction: (A) KOaq) + No,-(aq) KNoaq) 52. (B) Pb2+(aq) + 2cr(aq) Pbci2(s) (C) Pb*(aq)+2NO, (aq)+2K (aq) +2Cr(aq) PbCh(s) + 2K+(aq) + 2N03. (aq) (D) Pba (aq)+2NO, (ag)+2KCI(aq) PbClds) + 2K+(aq) + 2NO3- (aq) 53. What is the concentration of a 15.0 mL phosphoric acid solution that requires 10.0 mL of 0.200 M NaOH to react completely? H,POdaq) + 3NaOH(aq) Na,POdaq) + 3H2O(1) (A) 0.0444 M (C) 0.400 M (B) 0.100 M (D) 10.0 M 54. When 5.60 g CaF2 reacted with Molar Mass /g.mor excess H SO4, 1.73 g HF was CaF2 obtained. What is the percent HF yield? CaF2 + H2SO4 2HF + Caso, (A) 30.9% (B) 49.8% (C) 51.3% 78.07 20.01 (D) 60.3%

Explanation / Answer

50) 87 % of ethanol = 475 g

if it is 100 % then 475 g x 100 / 87 => 545.98 g

molar mass of ethanol = 46.07 g/mol

545.98 g / 46.07 g/mol => 11.85 mol

1 molecule of glucose produces 2 molecules of ethanol.

so 11.85 mol of ethanol could be produced by 11.85 / 2 => 5.93 mol of glucose

Mass of glucose = 5.93 mol x 180.2 g/mol => 1068 g => 1.07 x 103 g of glucose

51) 2 moles of ZnS requires 3 moles of O2

0.250 mol of O2 could react with 0.250 mol x 2/3 => 0.167 mol of ZnS

So remaining unreacted ZnS = 0.250 - 0.167 => 0.083 mol of ZnS.

52) Complete ionic equation

Pb^2+ (aq) + 2 NO3^- (aq) + 2 K^+ (aq) + 2 Cl^- (aq) ---> PbCl2 (s) + 2 K^+ (aq) + 2 NO3^- (aq)

So net ionic equation is ionic equation remains after cancel out common ions on both sides.

Answer : Pb^2+ (aq) + 2 Cl^- (aq) ---> PbCl2 (s) (option B).

53) moles of NaOH = 10 mL x 0.200 mol / 1000 mL => 0.002 mol

So moles of H3PO4 = 0.002 mol / 3 => 0.000666 moles

Concentration of H3PO4 = 0.000666 mole / 0.015 L => 0.0444 M

54) 5.60 g / 78.07 g/mol => 0.0717 moles

calculated yield of HF = 2 x 0.0717 mol x 20.01 g/mol => 2.87 g

% Yield = (actual yiled / calculated yield ) x 100

% Yield = ( 1.73 g / 2.87 g) x 100 => 60.3 %

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