The chief ore of zinc is the sulphide- ZnS- The ore is concentrated by floatation process and then heated in air- which converts the ZnS to ZnO- I- 2ZnS - 3O2 2ZnO - 2SO2The ZnO is then treated with dilute H2SO4- II- 2ZnO - 2H2SO4 2ZnSO4 - 2H2O Zn metal is produced on electrolysis- III- 2ZnSO4- 2H2O 2Zn-2H2SO4-O2Efficiencies of the processes have been indicated above the arrow mark- Mass of Zn will be obtained from an ore containing 225 kg of ZnS in kg is nearest integer value -

From the balanced chemical equations, it is observed that 2 mol ZnS ≡ 2 mol ZnO ≡ 2 mol ZnSO_4 ≡ 2 mol Zn The molar masses of Zn and ZnS a ...

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Standard XII

Chemistry

Stoichiometric Calculations

The chief ore...

Question

The chief ore of zinc is the sulphide, $Z n S$ . The ore is concentrated by floatation process and then heated in air, which converts the $Z n S$ to $Z n O$ .

$I .$ $2 Z n S + 3 O_{2} 90.6 % - -- \to 2 Z n O + 2 S O_{2}$
The $Z n O$ is then treated with dilute $H_{2} S O_{4}$ .

$I I .$ $2 Z n O + 2 H_{2} S O_{4} 100 % - -- \to 2 Z n S O_{4} + 2 H_{2} O$
$Z n$ metal is produced on electrolysis.

$I I I .$ $2 Z n S O_{4} + 2 H_{2} O 98.2 % - -- \to 2 Z n + 2 H_{2} S O_{4} + O_{2}$

Efficiencies of the processes have been indicated above the arrow mark. Mass of $Z n$ will be obtained from an ore containing $225$ kg of $Z n S$ in kg is (nearest integer value):

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Solution

From the balanced chemical equations, it is observed that
$2 m o l Z n S \equiv 2 m o l Z n O \equiv 2 m o l Z n S O_{4} \equiv 2 m o l Z n$
The molar masses of Zn and ZnS are 65 g/mol and 97 g/mol respectively.
Mass of Zn obtained will be $\frac{225000}{97} \times 65 = 134 g$

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Similar questions

The chief ore of

Z n

is the sulphide,

Z n S

. The ore is concentrated by froth floatation process and then heated in air to convert

Z n S

Z n O

2 Z n S + 3 O_{2} \to 2 Z n O + 2 S O_{2}

(

75 %

conversion)

Z n O + H_{2} S O_{4} \to Z n S O_{4} + H_{2} O

(

100 %

conversion )

2 Z n S O_{4} + 2 H_{2} O \to 2 Z n + 2 H_{2} S O_{4} + O_{2}

(

80 %

conversion)

The mass of

Z n

obtained from a sample of ore containing

292.2

kg of

Z n S

will be :

Q. The chief ore of

Z n

is the sulphide,

Z n S

. The ore is concentrated by froth floatation process and then heated in air to convert

Z n S

Z n O

2 Z n S + 3 O_{2} 80 % - - \to 2 Z n O + 2 S O_{2}

Z n O + H_{2} S O_{4} 100 % - -- \to Z n S O_{4} + H_{2} O

2 Z n S O_{4} + 2 H_{2} O 80 % - - \to 2 Z n + 2 H_{2} S O_{4} + O_{2}

The number of moles of

Z n S

required to produce 2 moles of

Z n

will be:

Q. The chief ore of

Z n

is its sulphide,

Z n S

. The ore is concentrated by froth floatation process and then heated in air to convert

Z n S

Z n O

. The following reactions take place in the extraction of

Z n

2ZnS+3O2→2ZnO+2SO2 (Yield: 62.5 \%)

ZnO+H2SO4→ZnSO4+H2O (Yield: 80 \%)

2ZnSO4+2H2O→2Zn+2H2SO4+O2 (Yield: 80\%)

The number of moles of

Z n S

required for producing

2

mol of

Z n

will be:

Q. The chief ore of

Z n

is its sulphide,

Z n S

. The ore is concentrated by froth floatation process and then heated in air to convert

Z n S

Z n O

. The following reactions take place in the extraction of

Z n

2 Z n S + 3 O_{2} \to 2 Z n O + 2 S O_{2} (Yield: 80 %)

Z n O + H_{2} S O_{4} \to Z n S O_{4} + H_{2} O (Yield: 100 %)

2 Z n S O_{4} + 2 H_{2} O \to 2 Z n + 2 H_{2} S O_{4} + O_{2} (Yield: 80%)

The number of moles of

Z n S

required for producing 2 mol of

Z n

will be:

Q. The chief ore of

Z n

is the sulphide,

Z n S

. The ore is concentrated by froth floatation process and then heated in air to convert

Z n S

Z n O

2 Z n S + 3 O_{2} 80 % - - \to 2 Z n O + 2 S O_{2}

Z n O + H_{2} S O_{4} 100 % - -- \to Z n S O_{4} + H_{2} O

2 Z n S O_{4} + 2 H_{2} O 80 % - - \to 2 Z n + 2 H_{2} S O_{4} + O_{2}

The number of moles of

Z n S

required to produce 2 moles of

Z n

will be:

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From the balanced chemical equations, it is observed that 2molZnS≡2molZnO≡2molZnSO4≡2molZn The molar masses of Zn and ZnS are 65 g/mol and 97 g/mol respectively.Mass of Zn obtained will be 22500097×65=134g

From the balanced chemical equations, it is observed that
$2 m o l Z n S \equiv 2 m o l Z n O \equiv 2 m o l Z n S O_{4} \equiv 2 m o l Z n$
The molar masses of Zn and ZnS are 65 g/mol and 97 g/mol respectively.
Mass of Zn obtained will be $\frac{225000}{97} \times 65 = 134 g$