Chapter 9, Problem 16A

(a)

Interpretation Introduction

Interpretation :

The amount of FeCO₃ and K₂SO₄ in mg must be calculated which will be obtained from 10 mg of FeSO₄. The mole ratio used for the conversion must also be clearly indicated.

Concept Introduction :

1 mol FeSO₄ (molar mass 152 g) produces 1 mol FeCO₃ (molar mass 116 g) and 1 mol K₂SO₄ (molar mass 174 g).

Answer to Problem 16A

Amount of FeCO₃ and K₂SO₄ in mg will be 7.63 mg and 11.45 mg respectively. Moles ratio used for both the cases is (1:1).

Explanation of Solution

The chemical reaction is represented as follows:

  ${\text{FeSO}}_{\text{4}}{\text{(aq)+K}}_{\text{2}}{\text{CO}}_{\text{3}}\text{(aq)}\to {\text{FeCO}}_{\text{3}}{\text{(s)+K}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}\text{.}$

As per the balanced chemical equation,

152 mg FeSO₄ produces 116 mg FeCO₃.

1 mg FeSO₄ produces $\frac{116}{152}$ mg FeCO₃.

10 mg FeSO₄ produces $\frac{\text{116}\times \text{10}}{\text{152}}\text{ mg =}\frac{\text{1160}}{\text{152}}\text{ mg = 7}\text{.63 mg}$ FeCO₃.

Here,

152 mg FeSO₄ produces 174 mg K₂SO₄.

1 mg FeSO₄ produces $\frac{\text{174}}{\text{152}}$ mg K₂SO₄.

10 mg FeSO₄ produces $\frac{\text{174}\times \text{10}}{\text{152}}\text{ mg =}\frac{\text{1740}}{\text{152}}\text{ mg = 11}\text{.45 mg}$ K₂SO₄.

b)

Interpretation Introduction

Interpretation :

The amount of CrCl₃ and Sn in mg must be calculated which will be obtained from 10 mg of Cr. The mole ratio used for the conversion must also be clearly indicated.

Concept Introduction :

4 mol CrCl₃ (molar mass 158 g) produces 4 mol Cr (molar mass 52 g) and 3 mol Sn (molar mass 119 g).

Answer to Problem 16A

Amount of CrCl₃ and Sn in mg will be 30.38 mg and 17.16 mg respectively. Mole ratio used for Cr to CrCl₃ is (1:1) and for Cr to Sn is (4:3).

Explanation of Solution

The reaction is represented as follows:

  ${\text{4Cr(s)+3SnCl}}_4\text{(aq)}\to {\text{4CrCl}}_{\text{3}}\text{(s)+3Sn(s)}\text{.}$

As per the balanced chemical equation,

52 mg Cr produces 158 mg CrCl₃.

1 mg Cr produces $\frac{158}{52}$ mg CrCl₃.

10 mg Cr produces $\frac{\text{158}\times \text{10}}{\text{52}}\text{ mg =}\frac{\text{1580}}{\text{52}}\text{ mg = 30}\text{.38 mg}$ CrCl₃.

Also,

  $4\times 52$ mg Cr produces $3\times 119$ mg Sn.

1 mg Cr produces $\frac{3\times 119}{4\times 52}$ mg Sn.

10 mg Cr produces $\text{=}\frac{\text{3×119×10}}{\text{4×52}}\text{ mg = }\frac{\text{3570}}{\text{208}}\text{ mg =17}\text{.16 mg}$ Sn.

c)

Interpretation Introduction

Interpretation :

The amount of Fe₂S₃ in mg must be calculated which will be obtained from 10 mg of S₈. The mole ratio used for the conversion must also be clearly indicated.

Concept Introduction :

3 mol S₈ (molar mass 256 g) produces 8 moles Fe₂S₃ (molar mass 208 g).

Answer to Problem 16A

Amount of Fe₂S₃ in mg will be 21.67 mg. Mole ratio used for S₈ to Fe₂S₃ is (3:8).

Explanation of Solution

The balanced chemical reaction is as follows:

  ${\text{16Fe(s)+3S}}_8\text{(s)}\to {\text{8Fe}}_{\text{2}}{\text{S}}_{\text{3}}\text{(s)}\text{.}$

As per the balanced chemical equation,

  $\text{3×256}$ mg S₈ produces $\text{8×208}$ mg Fe₂S₃.

1 mg S₈ produces $\frac{\text{8×208}}{\text{3×256}}$ mg Fe₂S₃.

10 mg S₈ produces $\frac{\text{8×208×10}}{\text{3×256}}\text{ mg =}\frac{\text{16640}}{\text{768}}\text{mg =21}\text{.67 mg}$ mg Fe₂S₃.

d)

Interpretation Introduction

Interpretation :

The amount of AgNO_3, H₂O and NO in mg must be calculated which will be obtained from 10 mg of HNO₃. The mole ratio used for the conversion must also be clearly indicated.

Concept Introduction :

4 moles of HNO₃ (molar mass 63 g) produces 3 moles AgNO₃ (molar mass 170 g)_, 2 moles of H₂O (molar mass 18 g) and one mole of NO (molar mass 30 g).

Answer to Problem 16A

Amount of AgNO_3, H₂O and NO in mg will be 202.38 mg, 1.43 mg and 1.19 mg respectively. Mole ratio used for HNO₃ to AgNO_3, H₂O and NO are (4:3), (4:1) and (4:2) respectively.

Explanation of Solution

The reaction is represented as follows:

  ${\text{3Ag(s)+4HNO}}_3\text{(aq)}\to {\text{3AgNO}}_{\text{3}}{\text{(aq)+2H}}_{\text{2}}\text{O(l)+NO(g)}\text{.}$

As per the balanced chemical equation,

  $\text{4×63}$ mg HNO₃ produces $3\times 170$ mg AgNO₃.

1 mg HNO₃ produces $\frac{3\times 170}{4\times 63}$ mg AgNO₃.

10 mg HNO₃ produces $\frac{\text{3×170×10}}{\text{4×63}}\text{ mg =}\frac{\text{5100}}{\text{252}}\text{ mg =202}\text{.38 mg}$ AgNO₃.

  $\text{4×63}$ mg HNO₃ produces $\text{2}\times \text{18}$ mg H₂O.

1 mg HNO₃ produces $\frac{2\times 18}{4\times 63}$ mg H₂O.

10 mg HNO₃ produces $\frac{\text{2}\times \text{18×10}}{\text{4×63}}\text{ mg =}\frac{\text{360}}{\text{252}}\text{ mg =1}\text{.43 mg}$ H₂O.

Also,

  $\text{4×63}$ mg HNO₃ produces 30 mg NO.

1 mg HNO₃ produces $\frac{30}{4\times 63}$ mg NO.

10 mg HNO₃ produces $\frac{\text{30×10}}{\text{4×63}}\text{ mg =}\frac{\text{300}}{\text{252}}\text{ mg =1}\text{.19 mg}$ NO.