Molar180.2 50. A particular yeast ferments glucose (Cali 0) and generates Mass/g

Question

Molar180.2 50. A particular yeast ferments glucose (Cali 0) and generates Mass/g mot a ethanol (CHsOH) with an 87.0% yield. what massof glucose is required to produce 475 g of ethanol by this process? CH OH 46.07 Cal1nOdaq) 2GHOH(aq) +2CO2(g) (A) 808 g (C) 1.07x 10'g (B) 929g (D) 2.14x 103 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 left? 51. are 2ZnS(s) + 302(g) 2ZnO(s) + 2S02(g) (B) 0.167 mol Zns (D) 0.250 mol O (A) 0.083 mol Zns (C) 0.125 mol O2 What is the net ionic equation for the reaction: Pb(NO3)2(aq) + 2KCI(aq) PbCl2(s) + 2KNOfaq)? (A) K.(aq) + NO3-(aq)- KNOs(aq) (B) Pb2+(aq) + 2C1(aq) PbCl2(s) (C) Pb2 (aq) + 2NOs (aq) + 2K (aq) + 2CI (ag) 52. -> PbCl2(s) + 2K'(aq) + 2NO, (aq) PbCl2(s) + 2K'(aq) + 2NO3 (aq) solution that requires 10.0 mL of 0.200 M NaOH to react (D) Pb2 (aq) +2NOs (aq) +2KCI(aq) 53. What is the concentration of a 15.0 mL phosphoric acid completely? HsPOa(aq) + 3 NaOH(aq) Na,PO4(aq) + 3H20(1) (A) 0.0444 M (C) 0.400 M (B) 0.100 M (D) 10.0 M 54. When 5.60 g CaF reacted with Molar Mass/g mol-i excess H2SO4. 1.73 g HF was CaF obtained. What is the percent yield? CaF2 + H2SO4 2HF + Caso, (A) 30.9% (B) 49.8% (C) 51.3% 78.07 20.01 HF (D) 60.3%

Explanation / Answer

50) write the balance reaction equation

C6H12O6 (aq) ------> 2 C2H5OH (aq) + 2CO2 (g)

1 mole of glucose produce 2 mole of ethanol and 2 mol of carbon dioxide

and percentage yeild of product from reactant is only 87.0 % so,

find the moles of ethanol = mass / molar mass = (475 g / 46.07 g/mol) = 10.31 moles

if 1 mole of glucose than ethanol will be = ( 1 x 2 ) 87 / 100 = 1.74 moles

so 10.31 moles ethanol required = (1 / 1.74) x 10.31 moles = 5.9253 moles

mass of glucose required = molar mass x moles = (180.2 g/mol x 5.9253 moles ) = 1067.73 g ~ 1.07x103 g

51) write the balance reation  

2ZnS (s) + 3O2 (g) --------> 2ZnO (s) + 2SO2 (g)  

2 mole of ZnS react with 3 mole of O2 and produce 2 mole of ZnO and 2 mole of SO2

we have 0.25 mole of ZnS and 0.25 mole of O2

but required 3 mole of O2 with 2 mole of ZnS

so O2 is limiting reagent

ammount of ZnS = (0.25 x 2/3 ) = 0.166 mole ZnS

excess amount of ZnS = 0.25 - 0.166 = 0.083 mol ZnS

52) write balance equation

Pb(NO3)2 (aq) + 2KCl (aq) ----> PbCl2(s) + 2KNO3 (aq)

[Pb]+2 + 2[Cl]- + 2[K]+ + 2[NO3]-  -------> 2K+ (aq) + 2NO3– (aq) + PbCl2 (s)

2Cl-(aq) + Pb2+(aq) ------> PbCl2 (s)

53) write balance reaction equation

H3PO4(aq) + 3NaOH(aq) ----> Na3PO4 (aq) + 3H2O(l)

1 mole of H3PO4 react with 3 mole of NaOH produce 1 mole of Na3PO4 and 3 mole of H2O so,

find moles of NaOH = molarity x volume = (0.2 M x 0.01 L ) = 0.002 moles

H3PO4 required = (0.002 moles / 3) = 0.000666 moles

molarity of H3PO4 = moles / volume = (0.000666 moles / 0.015 L ) = 0.0444 M

54) write balance reaction

CaF2 + H2SO4 ----> 2HF + CaSO4

1 mole of CaF2 react with 1 mole of H2SO4 produce 2 mole of Hf and 1 mole of CaSO4

moles of CaF2 = (mass / molar mass ) = (5.6 g / 78.07 g/mol) = 0.07173 moles

moles of HF = (1.73 g / 20.01 g/mol) = 0.086456 moles

1 mole of CaF2 produce 2 mole of HF

if 100 % yield than 0.14346 moles HF will be formed but here is only 0.086456 moles

% percentage yield = (0.086456 moles / 0.14346 moles) x 100 = 60.3 %  

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