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 determine the theoretical yield of PbCl2 using stoichiometry. The balanced chemical equation for the reaction is:
2 NaCl + Pb(NO3)2 -> 2 NaNO3 + PbCl2
From the equation, we can see that 2 moles of NaCl react with 1 mole of Pb(NO3)2, producing 1 mole of PbCl2. The molar masses of the relevant compounds are:
NaCl: 22.990 + 35.453 = 58.443 g/mol
Pb(NO3)2: 207.200 + 2(14.007 + 3(15.999)) = 331.200 g/mol
PbCl2: 207.200 + 2(35.453) = 278.106 g/mol
Using the masses given, we can calculate the moles of each reactant:
NaCl: 7.000 g / 58.443 g/mol = 0.1197 mol
Pb(NO3)2: 50.400 g / 331.200 g/mol = 0.1520 mol
Since 2 moles of NaCl react with 1 mole of Pb(NO3)2, the limiting reactant is NaCl. Therefore, the moles of PbCl2 produced is:
0.1197 mol NaCl x (1 mol PbCl2/2 mol NaCl) = 0.0598 mol PbCl2
Finally, we can calculate the theoretical yield of PbCl2:
0.0598 mol PbCl2 x 278.106 g/mol = 16.618 g PbCl2
The percent yield is then calculated by dividing the actual yield by the theoretical yield and multiplying by 100%:
percent yield = (12.798 g / 16.618 g) x 100% = 77.343%
Therefore, the answer is 77.343%.
To calculate the percent yield of PbCl2, we need to compare the actual yield to the theoretical yield.
First, we need to find the moles of NaCl and Pb(NO3)2 using their respective masses and molar masses:
Molar mass of NaCl (Na = 22.990 g/mol, Cl = 35.453 g/mol):
NaCl = 22.990 + 35.453 = 58.443 g/mol
Moles of NaCl:
moles = mass / molar mass = 7.000 g / 58.443 g/mol ≈ 0.1200 mol
Molar mass of Pb(NO3)2 (Pb = 207.200 g/mol, N = 14.007 g/mol, O = 15.999 g/mol):
Pb(NO3)2 = (207.200) + (2 × 14.007) + (6 × 15.999) = 331.200 g/mol
Moles of Pb(NO3)2:
moles = mass / molar mass = 50.400 g / 331.200 g/mol ≈ 0.1520 mol
Now, we need to determine the limiting reactant to find the theoretical yield of PbCl2.
The balanced chemical equation for the reaction is:
2NaCl + Pb(NO3)2 → 2PbCl2 + 2NaNO3
From the equation, we can see that the mole ratio between NaCl and Pb(NO3)2 is 2:1. Therefore, 0.1200 mol of NaCl will react with 0.0600 mol (0.1200 mol / 2) of Pb(NO3)2.
Now, calculate the theoretical yield of PbCl2 using the mole ratio:
Molar mass of PbCl2 (Pb = 207.200 g/mol, Cl = 35.453 g/mol):
PbCl2 = (207.200) + (2 × 35.453) = 278.106 g/mol
Theoretical moles of PbCl2:
moles = 0.0600 mol
Theoretical yield of PbCl2:
mass = moles × molar mass = 0.0600 mol × 278.106 g/mol ≈ 16.686 g
Now, calculate the percent yield:
% Yield = (Actual Yield / Theoretical Yield) × 100
% Yield = (12.798 g / 16.686 g) × 100 ≈ 76.672%
Therefore, the % yield of PbCl2 is approximately 76.672%.
None of the given options match the calculated result, so the closest answer would be 77.343%, although it is slightly different.