Question

Consider the reaction
\(4\text{HNO}_3(\text{l}) + 3\text{KCl}(\text{s}) \rightarrow \text{Cl}_2(\text{g}) + \text{NOCl}(\text{g}) + 2\text{H}_2\text{O}(\text{g}) + 3\text{KNO}_3(\text{s})\)
The amount of HNO₃ required to produce 110.0 g of KNO₃ is
(Given : Atomic masses of H, O, N and K are 1, 16, 14 and 39 respectively.)

• A. 32.2 g
• B. 69.4 g
• C. 91.5 g
• D. 162.5 g

Answer 1

The Correct Option is C

\(4\text{HNO}_3(\text{l}) + 3\text{KCl}(\text{s}) \rightarrow \text{Cl}_2(\text{g}) + \text{NOCl}(\text{g}) + 2\text{H}_2\text{O}(\text{g}) + 3\text{KNO}_3(\text{s})\)

\(\because 110 \, \text{g of KNO}_3 \Rightarrow \text{moles of KNO}_3 = \frac{110}{101} = 1.089 \, \text{mol}\)

As, \(4 \, \text{mol of HNO}_3 \text{ produces } 3 \, \text{mol of KNO}_3\).
\(\text{Hence, the moles of HNO}_3 \text{ required to produce } 1.089 \text{ moles of KNO}_3 =\)
\(=43×1.089=1.452 mol\)
\(\text{Hence, mass of HNO}_3 \text{ required} = 1.452 \times 63\)
\(≃ 91.5 g\)
So, the correct option is (C): \(91.5\ \text{g}\)