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Chemistry Question on Chemical Reactions

Consider the reaction
4HNO3(l)+3KCl(s)→Cl2(g)+NOCl(g)+2H2O(g)+3KNO3(s)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

91.5 g

Explanation

Solution

4HNO3(l)+3KCl(s)→Cl2(g)+NOCl(g)+2H2O(g)+3KNO3(s)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})

∵110 g of KNO3⇒moles of KNO3=110101=1.089 mol\because 110 \, \text{g of KNO}_3 \Rightarrow \text{moles of KNO}_3 = \frac{110}{101} = 1.089 \, \text{mol}

As, 4 mol of HNO3 produces 3 mol of KNO34 \, \text{mol of HNO}_3 \text{ produces } 3 \, \text{mol of KNO}_3.
Hence, the moles of HNO3 required to produce 1.089 moles of KNO3=\text{Hence, the moles of HNO}_3 \text{ required to produce } 1.089 \text{ moles of KNO}_3 =
=43×1.089=1.452mol=43×1.089=1.452 mol
Hence, mass of HNO3 required=1.452×63\text{Hence, mass of HNO}_3 \text{ required} = 1.452 \times 63
≃91.5g≃ 91.5 g
So, the correct option is (C): 91.5 g91.5\ \text{g}