The first step is to calculate the stoichiometric amount of Pb(NO3)2 required to react with 7.000 g of NaCl:
Moles of NaCl = (mass / molar mass) = 7.000 g / 58.443 g/mol = 0.1199 mol
Moles of Pb(NO3)2 = 2 * moles of NaCl = 0.2399 mol
Mass of Pb(NO3)2 = moles * molar mass = 0.2399 mol * 331.209 g/mol = 79.247 g
Since only 50.400 g of Pb(NO3)2 was used, it is limiting reagent and NaCl is in excess. Therefore, we need to calculate the theoretical yield of PbCl2 based on the amount of Pb(NO3)2 used:
Moles of Pb(NO3)2 = 50.400 g / 331.209 g/mol = 0.1520 mol
Moles of PbCl2 = moles of Pb(NO3)2 = 0.1520 mol
Mass of PbCl2 = moles * molar mass = 0.1520 mol * 278.104 g/mol = 42.147 g
Therefore, the theoretical yield of PbCl2 is 42.147 g. The percent yield is:
% yield = (actual yield / theoretical yield) * 100%
% yield = (12.798 g / 42.147 g) * 100% = 30.413%
Therefore, the correct answer is not provided in the options.
If the theoritical yield of PbCl2 is 16.547 g from the reaction of 7.000 g of NaCl with 50.400 g of Pb(NO3)2. Calculate the % yield of PbCl2 if 12.798 g of PbCl2 was actually obtained? (Atomic mass: Na-22.990 g/mol; 0-15.999 g/mol; Cl-35.453g/mol; Pb-207.200 g/mol; N-14.007).
100%
77.343%
38.343%
60.551%
1 answer
1 answer