Percent yield - 2 | Chemical Calculations

1) Determine the limiting reagent:

n_Pb(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
n_NaI = $\frac{{\mathrm{m}}_{\mathrm{NaI}}}{{\mathrm{M}}_{\mathrm{NaI}}}$ = $\frac{15.0}{149.9}$ = 1.00 x 10^-1 mol

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

2 n_Pb(NO3)2 = 1.51 x 10^-1 mol > n_NaI

NaI is the limiting reagent

2) Determine m_theoretical:

From the chemical equation: 2 moles of NaNO₃ are produced for 2 moles of NaI.
n_NaNO3 = n_NaI
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 = m_obtained / m_theoretical x 100 = $\frac{7.50}{8.51}$ x 100 = 88.1 %