Consider the reaction

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

The amount of HNO₃ required to produce 110.0g of KNO₃ is

(Given: Atomic masses of H, O, N and K are 1, 16, 14 and 39, respectively)

Option 1 -

32.2g

Option 2 -

69.4g

Option 3 -

91.5g

Option 4 -

162.5g

4 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

$4 H N O I_{3 (l)} + 3 K C l_{(s)} \to_{}^{} C l_{2 (g)} + N O C l_{(g)} + 2 H_{2} O (g) + 3 K N O_{3} (g)$

$↓$

4 moles of HNO₃ produced 3 mol of KNO₃

Here mole of produced KNO₃ = $\frac{110}{101}$

If 3 mol of KNO₃ produced by 4 moles of HNO₃

$∴$ 1 mole of KNO₃ produced by $\frac{4}{3}$ moles of HNO₃

and $\frac{110}{101}$ mole of KNO₃ produced by $\frac{4 \times 110}{3 \times 101}$ moles of HNO₃ = 1.45 mole of HNO₃

Hence mass of HNO₃= mole × mol.wt = 145 × 63 = 91.48 $\approx$ 91.5gm

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