# Exercise 8 | Chemical Calculations

15.0 g of copper(II) chloride react with 20.0 g of sodium nitrate to form copper(II) nitrate and sodium chloride.

1) Write the chemical equation of this reaction

2) What is the limiting reagent?

3) How many grams of sodium chloride is formed?

4) What is the percent yield of this reaction if 10.0 grams of sodium chloride are formed?

1) CuCl2 + 2 NaNO3 → Cu(NO3)2 + 2 NaCl

2) nCuCl2 = $\frac{{\mathrm{m}}_{\mathrm{CuCl}2}}{{\mathrm{M}}_{\mathrm{CuCl}2}}$ = $\frac{15.0}{134.45}$ = 1.11 x 10-1 mol
nNaNO3 = $\frac{{\mathrm{m}}_{\mathrm{NaNO}3}}{{\mathrm{M}}_{\mathrm{NaNO}3}}$ = $\frac{20.0}{85.0}$ = 2.35 x 10-1 mol

From the chemical equation: 2 moles of NaNO3 react with 1 mole of CuCl2.
2 nCuCl2 = 2.22 10-1 mol < nNaNO3

CuCl2 is the limiting reagent

3) From the chemical equation: 2 moles of NaCl are produced for each mole of CuCl2.
nNaCl = 2 nCuCl2
nNaCl = 2 x $\frac{{\mathrm{m}}_{\mathrm{CuCl}2}}{{\mathrm{M}}_{\mathrm{CuCl}2}}$ = 2 x $\frac{15.0}{134.45}$ = 2.23 x 10-1 mol
mNaCl = nNaCl x MNaCl = 2.23 x 10-1 x 58.4 = 13.0 g

4) % yield = mobtained / mtheoretical x 100 =  x 100 = 76.9 %