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).

Answer 1

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.