Mole Concept and Stoichiometry — Question 4

$\begin{matrix} \text{KNO}_3 & + &\text{H}_2\text{SO}_4 & \longrightarrow & \text{KHSO}_4 & + \text{HNO}_3 \\ 39 + 14 + 3(16) & & & & & 1 + 14 + 3(16) \\ = 39 + 14 + 48 & & & & & = 1 + 14 + 48 \\ = 101 \text{ g} & & & & & = 63 \text{ g} \end{matrix}$

63 g of HNO₃ is formed by 101 g of KNO₃

∴ 126000 g of HNO₃ is formed by $\dfrac{101}{63}$ x 126000
= 202000 g = 202 kg

Similarly,

$\begin{matrix} \text{NaNO}_3 & + &\text{H}_2\text{SO}_4 & \longrightarrow & \text{NaHSO}_4 & + \text{HNO}_3 \\ 23 + 14 + 3(16) & & & & & 1 + 14 + 3(16) \\ = 23 + 14 + 48 & & & & & = 1 + 14 + 48 \\ = 85 \text{ g} & & & & & = 63 \text{ g} \end{matrix}$

63 g of HNO₃ is formed by 85 g of NaNO₃

∴ 126000 g of HNO₃ is formed by $\dfrac{85}{63}$ x 126000
= 170000 g = 170 kg

So, a smaller mass of NaNO₃ is required.