ZnSO4-7H2O A B O2 -SO2 -ZnO- A and B respectively are-

ZnSO_4.7H_2O ZnSO_4.6H_2O ZnSO_4.H_2O O_2 +SO_2 +ZnO ZnSO_4 (Anhydrous) This reaction is self explanatory. As ZnSO_4.7H_2O is heated above 39 degree Cel ...

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Standard XII

Chemistry

Zinc II Sulphate

ZnSO4.7H2O A ...

Question

$Z n S O_{4} .7 H_{2} O A b o v e 39^{\circ} C - ------ \to B e l o w 70^{\circ} C A A b o v e 70^{\circ} C - ------ \to B A b o v e 280^{\circ} C - ------ \to O_{2} + S O_{2} + Z n O$ . A and B respectively are:

Z n S O_{4} . H_{2} O & Z n S O_{4} .6 H_{2} O

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Z n S O_{4} .2 H_{2} O & Z n S O_{4} .6 H_{2} O

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Z n S O_{4} .6 H_{2} O & Z n S O_{4} .2 H_{2} O

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Z n S O_{4} .6 H_{2} O & Z n S O_{4} . H_{2} O

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Solution

The correct option is D $Z n S O_{4} .6 H_{2} O & Z n S O_{4} . H_{2} O$
$Z n S O_{4} .7 H_{2} O A b o v e 39^{\circ} C - ------ \to B e l o w 70^{\circ} C Z n S O_{4} .6 H_{2} O A b o v e 70^{\circ} C - ------ \to Z n S O_{4} . H_{2} O A b o v e 280^{\circ} C - ------ \to O_{2} + S O_{2} + Z n O 800^{\circ} C \leftarrow -- - \begin{matrix} Z n S O_{4} (A n h y d r o u s) \end{matrix}$

This reaction is self-explanatory. As $Z n S O_{4} .7 H_{2} O$ is heated above 39 degree Celsius and below 70 degree Celsius, it loses one water molecule and further when $Z n S O_{4} .6 H_{2} O$ is heated above 70 degree Celsius, it loses 5 water molecules. Therefore, A is $Z n S O_{4} .6 H_{2} O$ and B is $Z n S O_{4} . H_{2} O$ .

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Z n S O_{4} .7 H_{2} O A b o v e 39^{\circ} C - ------ \to B e l o w 70^{\circ} C A A b o v e 70^{\circ} C - ------ \to B A b o v e 280^{\circ} C - ------ \to O_{2} + S O_{2} + Z n O

. A and B respectively are:

Z n S O_{4} .7 H_{2} O A b o v e 39^{\circ} C - ------ \to B e l o w 70^{\circ} C A A b o v e 70^{\circ} C - ------ \to B A b o v e 280^{\circ} C - ------ \to O_{2} + S O_{2} + Z n O

. A and B respectively are:

Q. Following method of extracting

Z n

is based on thermodynamics
A :

2 Z n S + 3 O_{2} ⟶ 2 Z n O + 2 S O_{2}

B :

Z n O + C ⟶ Z n + C O

△ G_{f}^{o}

(standard free energies of formation, in

k J m o l^{- 1}

) of

Z n S = - 205.4, Z n O = - 318.0, S O_{2} = - 300.4

and of

C O = - 137.3

Free energy changes of the above reaction A and B (respectively) will be :

Q. In the following reactions,

Z n O

is respectively acting as a/an:
A)

Z n O + N a_{2} O \to N a_{2} Z n O_{2}

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

Q. In the following reactions,

Z n O

is respectively acting as a/an:
(a)

Z n O + {N a}_{2} O ⟶ {N a}_{2} Z n O_{2}

(b)

Z n O + C O_{2} ⟶ Z n C O_{3}

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ZnSO4.7H2OAbove 39∘C−−−−−−−→Below 70∘CAAbove 70∘C−−−−−−−→BAbove 280∘C−−−−−−−→O2+SO2+ZnO. A and B respectively are:

$Z n S O_{4} .7 H_{2} O A b o v e 39^{\circ} C - ------ \to B e l o w 70^{\circ} C A A b o v e 70^{\circ} C - ------ \to B A b o v e 280^{\circ} C - ------ \to O_{2} + S O_{2} + Z n O$ . A and B respectively are: