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.)
\(4\text{HNO}_3(\text{l}) + 3\text{KCl}(\text{s}) \rightarrow \text{Cl}_2(\text{g}) + \text{NOCl}(\text{g}) + 2\text{H}_2\text{O}(\text{g}) + 3\text{KNO}_3(\text{s})\)
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.)
- 32.2 g
- 69.4 g
- 91.5 g
- 162.5 g
The Correct Option is C
Approach Solution - 1
To determine the amount of \(\text{HNO}_3\) required to produce 110.0 g of \(\text{KNO}_3\), we'll follow these steps:
- First, calculate the molar mass of \(\text{KNO}_3\).
| Element | Atomic Mass | Number of Atoms | Total Mass |
|---|---|---|---|
| K | 39 | 1 | 39 |
| N | 14 | 1 | 14 |
| O | 16 | 3 | 48 |
| Total Molar Mass of \(\text{KNO}_3\) | 101 g/mol | ||
- Calculate the number of moles of \(\text{KNO}_3\) in 110.0 g.
\(\text{Moles of KNO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{110.0 \, \text{g}}{101 \, \text{g/mol}} \approx 1.089 \, \text{mol}\)
- Use the stoichiometry of the reaction to determine the moles of \(\text{HNO}_3\) required.
According to the balanced chemical equation:
\(4\text{HNO}_3 + 3\text{KCl} \rightarrow \ldots + 3\text{KNO}_3 \\)
3 moles of \(\text{KNO}_3\) require 4 moles of \(\text{HNO}_3\).
\(\text{Moles of HNO}_3 = \frac{4}{3} \times \text{Moles of KNO}_3 = \frac{4}{3} \times 1.089 \approx 1.452 \, \text{mol}\)
- Calculate the mass of \(\text{HNO}_3\) required.
The molar mass of \(\text{HNO}_3\) is 63 g/mol.
\(\text{Mass of HNO}_3 = \text{moles} \times \text{molar mass} = 1.452 \, \text{mol} \times 63 \, \text{g/mol} \approx 91.4 \, \text{g}\)
The closest option to our calculated mass is 91.5 g. Therefore, the correct answer is 91.5 g.
Hence, the amount of \(\text{HNO}_3\) required is 91.5 g.
Approach Solution -2
\(4\text{HNO}_3(\text{l}) + 3\text{KCl}(\text{s}) \rightarrow \text{Cl}_2(\text{g}) + \text{NOCl}(\text{g}) + 2\text{H}_2\text{O}(\text{g}) + 3\text{KNO}_3(\text{s})\)
\(\because 110 \, \text{g of KNO}_3 \Rightarrow \text{moles of KNO}_3 = \frac{110}{101} = 1.089 \, \text{mol}\)
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): \(91.5\ \text{g}\)
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