Chem. 110.002 Quiz # 2 DUE 10/23/2017 SHOw YoUR WORK VERY, VERY CLEARLY 1. Balan
ID: 550611 • Letter: C
Question
Chem. 110.002 Quiz # 2 DUE 10/23/2017 SHOw YoUR WORK VERY, VERY CLEARLY 1. Balance the following equation:s a) PbO3 + HCl H2O + PbCl2 + Cl2 b) C2H6S+Og CO2+H20 +SO2 c) La2(COn3 + H2SO4 La2(SO4)3 + H2O + CO2 2. Caustic soda (NaOH) is prepared commercially by passing and electric current through a concentrated solution of salt water: 2NaCl(aq) + 2H2O 2NaOH(aq) + H2(g) + Cl2 What is the theoretical yield of caustic soda if 100 kg of sodium chloride is electrolyzed? a) b) What is the percent yield if the electrolysis produces 55 kg of caustic soda? 3. Calcium carbonate dissolves in hydrochloric acid according to the equation: CaCO3(s) + 2HCl(aq) CaCl2(aq) + H2O + CO2 How many grams of calcium chloride will form if 50.00 g of CaCOs is mixed with 0.500 mole of HCI? 4. A 0.1888-g sample of a hydrocarbon produces 0.6260 g CO2 and 0.1602 g H2O in combustion analysis. Its molecular mass is found to be 106 g/mol. For this hydrocarbon, determine a) its empirical formula and b) its molecular formula. What is the molarity of the following solutes? a) sucrose (C12H2011) if 100.0 g are dissolved in 500 mL of water, b) urea (CO(NH2)2 if 87.8 mg of the 98.7 % pure solid are dissolved in 5.00 mL of aqueous solution. 5. 6. Twenty-five mL of 0.388 M solution of Na2SO4 is mixed with 35.3 mL of 0.229 M Na SO4, What is molarity of the resulting solution. Assume that the volumes are additive.Explanation / Answer
1.
a. PbO2 + 4HCl --------> 2H2O + PbCl2 + Cl2
b. 2C2H6S + 9O2 ---------> 4CO2 + 6H2O+ 2SO2
c. La2(CO3)3 + 3H2So4 ------> La2(SO4)3 + 3H2O + 3CO2
2.
2NaCl + 2H2O ---------> 2NaOH + H2 + Cl2
2 moles of NaCl react with H2O to gives 2 moles of NaOH
2*58.5g of NaCl react with H2O to gives 2*40g of NaOH
100Kg of NaCl react with H2O to gives = 2*40g*100kg/2*58.5 = 68.4Kg
theoritical yield of naOH = 68.4Kg
percentage yield = actual yield*100/theoritical yield
= 55*100/68.4 = 80.4%
3. CaCo3(s) + 2HCl ------> CaCl2 + H2O + Co2
no of moles of CaCo3 = w/g.M.Wt
= 50/100 = 0.5 moles
1moles of CaCO3 react with 2 moles of HCl
0.5 moles of CaCo3 react with = 2*0.5/1 = 1 moles of HCl
HCl is limiting reactant
2 moles of HCl react with CaCo3 to gives 1 moles of CaCl2
0.5 moles of Hcl react with CaCo3 to gives = 0.5*1/2 = 0.25 moles of CaCl2
mass of CaCl2 = no of moles * gram molar mass
= 0.25*111 = 27.75g
4. C% = 12*wt of Co2*100/44*wt of hydrocarbon
= 12*0.626*100/44*0.1888 = 90.43%
H% = 2*wt of H2O *100/18*wt of hydrocarbon
= 2*0.1602*100/18*0.1888 = 9.43%
Element % A.Wt relative number simple ratio simplest ratio
C 90.43 12 90.43/12 = 7.54 7.54/7.54 = 1 1*4 = 4
H 9.43 1 9.43/1 = 9.43 9.43/7.54 = 1.25 1.25*4 =5
Empirical formula = C4H5
E.F.Wt = 53g/mole
molecular formula = (empirical formula)n
n = M.Wt/E.F.Wt
= 106/53 = 2
molecular fomula = (C4H5)2 = C8H10
=
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