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%
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.
To find the percent yield of PbCl2, we need to compare the actual yield with the theoretical yield.
First, let's calculate the theoretical yield of PbCl2:
1. Calculate the number of moles of NaCl:
moles of NaCl = mass of NaCl / molar mass of NaCl
= 7.000 g / 22.990 g/mol
= 0.304 mol
2. Calculate the number of moles of Pb(NO3)2:
moles of Pb(NO3)2 = mass of Pb(NO3)2 / molar mass of Pb(NO3)2
= 50.400 g / 331.209 g/mol
= 0.152 mol
3. Determine the limiting reactant:
The limiting reactant is the one that produces the least amount of the desired product. It is determined by comparing the mole ratio between the reactants and the product.
From the balanced chemical equation:
2 NaCl + Pb(NO3)2 → 2 PbCl2 + 2 NaNO3
The mole ratio of NaCl to Pb(NO3)2 is 2:1.
Since we have 0.304 moles of NaCl and 0.152 moles of Pb(NO3)2, the limiting reactant is Pb(NO3)2.
4. Calculate the theoretical yield of PbCl2:
The mole ratio of Pb(NO3)2 to PbCl2 is 1:1.
Therefore, the moles of Pb(NO3)2 (0.152 mol) will produce the same moles of PbCl2.
mass of PbCl2 = moles of PbCl2 × molar mass of PbCl2
= 0.152 mol × 278.102 g/mol
= 42.241 g
Now, we can calculate the percent yield:
Percent yield = (actual yield / theoretical yield) × 100
= (12.798 g / 42.241 g) × 100
= 30.3552%
Therefore, the percent yield of PbCl2 is approximately 30.3552%, which can be rounded to 30.355% or 30.4% (depending on the level of precision required). However, none of the provided answer choices match this result.