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): \(91.5\ \text{g}\)
Top JEE Main Chemistry Questions
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- At equilibrium the concentrations of and in a sealed vessel at . What will be for the reaction
- Find out the major products from the following reactions
Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
- The correct order of atomic radii is:
Top JEE Main Chemical Reactions Questions
- The products A and B formed in the following reaction scheme are respectively
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: H2Te is more acidic than H2S.
Reason R: Bond dissociation enthalpy of H2Te is lower than H2S.
In light of the above statements, choose the most appropriate from the options given below:
- Which of the following reactions are disproportionation reactions?
\(\text{(A) } \text{Cu}^+ \rightarrow \text{Cu}^{2+} + \text{Cu}\)
\(\text{(B) } 3\text{MnO}_4^{2-} + 4\text{H}^+ \rightarrow 2\text{MnO}_4^- + \text{MnO}_2 + 2\text{H}_2\text{O}\)
\(\text{(C) } 2\text{KMnO}_4 \rightarrow \text{K}_2\text{MnO}_4 + \text{MnO}_2 + \text{O}_2\)
\(\text{(D) } 2\text{MnO}_4^- + 3\text{Mn}^{2+} + 2\text{H}_2\text{O} \rightarrow 5\text{MnO}_2 + 4\text{H}^+\)
Choose the correct answer from the options given below. - Products for the below reaction are:
- Major product of the following reaction is
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.