# Exercise 9 | Chemical Calculations

Here is a chemical reaction:

Pb(NO_{3})_{2} + 2 NaI → PbI_{2} + 2 NaNO_{3}

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:__

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(NO_{3})_{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 NaNO_{3} 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 %