Baking soda (NaHCO $_{3}$ ) can be made in large quantities by the following reaction:
$\mathrm{NaCl}(a q)+\mathrm{NH}_{3}(a q)+\mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow$.$$\mathrm{NaHCO}_{3}(s)+\mathrm{NH}_{4} \mathrm{Cl}(a q)$$
If $10.0 \mathrm{g}$ of $\mathrm{NaCl}$ reacts with excesses of the other reactants and $4.2 \mathrm{g}$ of $\mathrm{NaHCO}_{3}$ is isolated, what is the percent yield of the reaction?