Limiting reactant | Chemical Calculations

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

2) n_CuCl2 = $\frac{{\mathrm{m}}_{\mathrm{CuCl}2}}{{\mathrm{M}}_{\mathrm{CuCl}2}}$ = $\frac{15.0}{134.45}$ = 1.11 x 10^-1 mol
n_NaNO3 = $\frac{{\mathrm{m}}_{\mathrm{NaNO}3}}{{\mathrm{M}}_{\mathrm{NaNO}3}}$ = $\frac{20.0}{85.0}$ = 2.35 x 10^-1 mol

From the chemical equation: 2 moles of NaNO₃ react with 1 mole of CuCl₂.
2 n_CuCl2 = 2.22 10^-1 mol < n_NaNO3

CuCl2 is the limiting reagent

3) From the chemical equation: 2 moles of NaCl are produced for each mole of CuCl₂.
n_NaCl = 2 n_CuCl2
n_NaCl = 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
m_NaCl = n_NaCl x M_NaCl = 2.23 x 10^-1 x 58.4 = 13.0 g

4) % yield = m_obtained / m_theoretical x 100 =  x 100 = 76.9 %