# Percent yield - 2 | Chemical Calculations

Here is a chemical reaction:

Pb(NO3)2 + 2 NaI → PbI2 + 2 NaNO3

What is the percent yield of the reaction if 7.50 g of sodium nitrate are formed?

1) Determine the limiting reagent:

nPb(NO3)2 = $\frac{{\mathrm{m}}_{\mathrm{Pb}\left(\mathrm{NO}3\right)2}}{{\mathrm{M}}_{\mathrm{Pb}\left(\mathrm{NO}3\right)2}}$ = $\frac{25.0}{331.2}$ = 7.55 x 10-2 mol
nNaI = $\frac{{\mathrm{m}}_{\mathrm{NaI}}}{{\mathrm{M}}_{\mathrm{NaI}}}$ = $\frac{15.0}{149.9}$ = 1.00 x 10-1 mol

From the chemical equation: 2 moles of NaI react with 1 mole of Pb(NO3)2.

2 nPb(NO3)2 = 1.51 x 10-1 mol > nNaI

NaI is the limiting reagent

2) Determine mtheoretical:

From the chemical equation: 2 moles of NaNO3 are produced for 2 moles of NaI.
nNaNO3 = nNaI
n NaNO3 = $\frac{{\mathrm{m}}_{\mathrm{NaI}}}{{\mathrm{M}}_{\mathrm{NaI}}}$ = $\frac{15.0}{149.9}$ = 1.00 x 10-1 mol
m NaNO3 = n NaNO3 x M NaNO3 = 1.00 x 10-1 x 85.0 = 8.51 g

3) % yield = mobtained / mtheoretical x 100 = $\frac{7.50}{8.51}$ x 100 = 88.1 %