Question

A student carefully adds 7.50g of NaHCO3 to an excess of HNO3. The actual yield of this reaction is 3.25g of CO2. What is the % yield?

Answer 1

The reaction for this process is next:

$$\mathrm{NaHCO_3} + \mathrm{HNO_3} \rightarrow \mathrm{NaNO_3} + \mathrm{CO_2} + \mathrm{H_2O}$$

The mole ratio of NaHCO3 and CO2 is 1: 1, so if amount is $n = \mathrm{m/Mw}$, for NaHCO3 it is:

$$n = \mathrm{m/Mw}$$ $$n = 7.50 / 84 = 0.0893 \mathrm{~mol}$$

If mole ratio of NaHCO3 and CO2 is 1: 1, the amount of CO2 is the same, and mass of it is:

$\mathrm{m = n*Mw = 0.0893*44 = 3.93g}$. It is a theoretical mass. Real mass is $3.25\mathrm{g}$.

$\%$ yield = real mass/ theoretical mass * $100\% = 3.25 / 3.93 * 100\% = 82.7\%$