Exercise 9 | Chemical Calculations

Here is a chemical reaction:

Pb(NO3)2 + 2 NaI → PbI2 + 2 NaNO3

We start with 25.0 g of lead(II) nitrate and 15.0 g of sodium iodide.

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 = mPb(NO3)2MPb(NO3)2 = 25.0331.2 = 7.55 x 10-2 mol
nNaI = mNaIMNaI = 15.0149.9 = 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 = mNaIMNaI = 15.0149.9 = 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 = 7.508.51 x 100 = 88.1 %