If the theoretical yield of PbCl2 is 16.547g from the reaction of 7.000g of NaCl with 50.400g of Pb(NO3)2. Calculate the % yield of PbCl2 if 12.798g of PbCl2 was actually obtained? (Atomic mass: Na-22.990 g/mol; Cl-35.453g/mol; 0-15.999g/mol; Pb-207.200 g/mol; N-14.007).
First, we need to write the balanced chemical equation for the reaction:
2 NaCl + Pb(NO3)2 → 2 NaNO3 + PbCl2
Next, we need to determine the limiting reagent to see how much PbCl2 can be formed.
To do this, we can convert the masses of NaCl and Pb(NO3)2 to moles:
moles of NaCl = 7.000g / 22.990 g/mol = 0.304 mol
moles of Pb(NO3)2 = 50.400g / 331.200 g/mol = 0.152 mol
Now, we can use the coefficients in the balanced equation to see which reactant runs out first:
NaCl: 2 moles react with 1 mole of Pb(NO3)2
0.304 mol NaCl x (1 mol Pb(NO3)2 / 2 mol NaCl) = 0.152 mol Pb(NO3)2
Pb(NO3)2: 1 mole reacts with 2 moles of NaCl
0.152 mol Pb(NO3)2 x (2 mol NaCl / 1 mol Pb(NO3)2) = 0.304 mol NaCl
Since we have equal amounts of both reactants, NaCl is the limiting reagent.
Now, we can use the mole ratio from the balanced equation to calculate the theoretical yield of PbCl2:
1 mol PbCl2 is formed from 2 moles of NaCl
0.304 mol NaCl x (1 mol PbCl2 / 2 mol NaCl) x (207.200 g/mol PbCl2) = 10.010g PbCl2
Therefore, the percent yield of PbCl2 is:
(actual yield / theoretical yield) x 100%
(12.798g / 10.010g) x 100% = 127.6%
This is not a reasonable percent yield because it is greater than 100%. It's possible that some impurities or water were present in the final product, leading to a higher measured mass.
To calculate the percent yield of PbCl2, we need to compare the actual yield (12.798g) with the theoretical yield (16.547g).
Step 1: Convert the mass of NaCl to moles.
Given:
Mass of NaCl = 7.000g
Molar mass of NaCl = 22.990g/mol
Moles of NaCl = mass of NaCl / molar mass of NaCl
= 7.000g / 22.990g/mol
≈ 0.304 mol
Step 2: Convert the mass of Pb(NO3)2 to moles.
Given:
Mass of Pb(NO3)2 = 50.400g
Molar mass of Pb(NO3)2 = (207.200g/mol) + (14.007g/mol * 3) + (15.999g/mol * 6)
Moles of Pb(NO3)2 = mass of Pb(NO3)2 / molar mass of Pb(NO3)2
= 50.400g / [(207.200g/mol) + (14.007g/mol * 3) + (15.999g/mol * 6)]
≈ 0.156 mol
Step 3: Determine the limiting reactant.
To determine the limiting reactant, we compare the moles of NaCl and Pb(NO3)2.
The balanced equation for the reaction is:
2 NaCl + Pb(NO3)2 → 2 NaNO3 + PbCl2
From the balanced equation, we can see that the ratio of NaCl to Pb(NO3)2 is 2:1.
Comparing the moles of NaCl and Pb(NO3)2:
Moles of NaCl: 0.304 mol
Moles of Pb(NO3)2: 0.156 mol
Based on the ratio, we can see that NaCl is in excess.
Step 4: Calculate the theoretical yield of PbCl2.
To calculate the theoretical yield of PbCl2, we need to consider the limiting reactant, which is Pb(NO3)2.
From the balanced equation, we can see that the ratio of Pb(NO3)2 to PbCl2 is 1:1.
Moles of PbCl2 = Moles of Pb(NO3)2
= 0.156 mol
Mass of PbCl2 = Moles of PbCl2 * Molar mass of PbCl2
= 0.156 mol * 278.643g/mol
≈ 43.440g
Step 5: Calculate the percent yield.
Percent yield = (Actual yield / Theoretical yield) * 100
Actual yield = 12.798g
Theoretical yield = 43.440g
Percent yield = (12.798g / 43.440g) * 100
≈ 29.50%
Therefore, the percent yield of PbCl2 is approximately 29.50%.