Section -A

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

3 Views | Posted 3 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

3 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

<math><mrow><mn>4</mn><mi>H</mi><mi>N</mi><mi>O</mi><msub><mrow><mi>I</mi></mrow><mrow><mn>3</mn><mrow><mo>(</mo><mrow><mi mathvariant="script">l</mi></mrow><mo>)</mo></mrow></mrow></msub><mo>+</mo><mn>3</mn><mi>K</mi><mi>C</mi><msub><mrow><mi>l</mi></mrow><mrow><mrow><mo>(</mo><mrow><mi>s</mi></mrow><mo>)</mo></mrow></mrow></msub><munderover><mo>→</mo><mrow></mrow><mrow></mrow></munderover><mi>C</mi><msub><mrow><mi>l</mi></mrow><mrow><mn>2</mn><mrow><mo>(</mo><mrow><mi>g</mi></mrow><mo>)</mo></mrow></mrow></msub><mo>+</mo><mi>N</mi><mi>O</mi><mi>C</mi><msub><mrow><mi>l</mi></mrow><mrow><mrow><mo>(</mo><mrow><mi>g</mi></mrow><mo>)</mo></mrow></mrow></msub><mo>+</mo><mn>2</mn><msub><mrow><mi>H</mi></mrow><mrow><mn>2</mn></mrow></msub><mi>O</mi><mrow><mo>(</mo><mrow><mi>g</mi></mrow><mo>)</mo></mrow><mo>+</mo><mn>3</mn><mi>K</mi><mi>N</mi><msub><mrow><mi>O</mi></mrow><mrow><mn>3</mn></mrow></msub><mrow><mo>(</mo><mrow><mi>g</mi></mrow><mo>)</mo></mrow></mrow></math> <math> <mrow> <mo>↓</mo> </mrow> </math>              4 moles of HNO3 produced 3 mol of KNO3Here mole of produced KNO3 = <math> <mrow> <mfrac> <mrow> <mn>1</mn> <mn>1</mn> <mn>0</mn> </mrow> <mrow> <mn>1</mn> <mn>0</mn> <mn>1</mn> </mrow> </mfrac> </mrow> </math>  If 3 mol of KNO3 produced by 4 moles of HNO3 <math> <mrow> <mo>∴</mo> </mrow> </math>  1 mole of KNO3 produced by  <math> <mrow> <mfrac> <mrow> <mn>4</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math> moles of HNO3and  <math> <mrow> <mfrac> <mrow> <mn>1</mn> <mn>1</mn> <mn>0</mn> </mrow> <mrow> <mn>1</mn> <mn>0</mn> <mn>1</mn> </mrow> </mfrac> </mrow> </math> mole of KNO3 produced by <math> <mrow> <mfrac> <mrow> <mn>4</mn> <mo>×</mo> <mn>1</mn> <mn>1</mn> <mn>0</mn> </mrow> <mrow> <mn>3</mn> <mo>×</mo> <mn>1</mn> <mn>0</mn> <mn>1</mn> </mrow> </mfrac> </mrow> </math>  moles of HNO3 = 1.45 mole of HNO3Hence mass of HNO3 = mole × mol.wt = 145 × 63 = 91.48 <math> <mrow> <mo>≈</mo> </mrow> </math> 91.5gm

0 Upvotes 0 Downvotes

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
678k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Section -A

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)

1 Answer

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

Section -A 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)

1 Answer

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question

Section -A

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)