In the Solvay process for producing sodium carbonate (Na₂CO₃), the following reactions occur in sequence

${\mathrm{NH}}_3+{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}⟶{\mathrm{NH}}_4{\mathrm{HCO}}_3$

${\mathrm{NH}}_4{\mathrm{HCO}}_3+\mathrm{NaCl}⟶{\mathrm{NH}}_4\mathrm{Cl}+{\mathrm{NaHCO}}_3$

$2{\mathrm{NaHCO}}_3\stackrel{\mathrm{\Delta }}{⟶}{\mathrm{Na}}_2{\mathrm{CO}}_3+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$

How much of Na₂CO₃ would be produced per kg of NH₃ used if the process were 100% efficient?

Given reaction is not balanced
Based on stoichiometry
1 mol Na₂CO₃ is from 2 mol NaHCO₃
2 mol NaHCO, is from 2 mol NH₄HCO₃
2 mol NH₄HCO₃ is from 2 mol NHI
Thus, 1 mol Na₂CO₃ is from = 2 mol NH₃
106 g NarCO₃ is from = 2 × 17 g NH₃
34 g NH₃ gives = 106 g Na₂CO₃ 

$\text{ Thus, }\begin{array}{r}1000\mathrm{g}{\mathrm{NH}}_3\text{ gives }=\frac{106\times 1000}{34}\mathrm{g}{\mathrm{Na}}_2{\mathrm{CO}}_3\\ =3118\mathrm{g}{\mathrm{Na}}_2{\mathrm{CO}}_3\\ =3.118\mathrm{kg}{\mathrm{Na}}_2{\mathrm{CO}}_3\end{array}$