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Action of Water on Calcium Carbide

In the follow...

Question

In the following redox reaction, number of moles of $C u {(O H)}_{2}$ reduced by one mole of $N_{2} H_{4}$ is:
$C u {(O H)}_{2} (s) + N_{2} H_{4} (a q) ⟶ C u (s) + N_{2} (g)$

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Solution

The balanced reaction is as follows:
$2 C u {(O H)}_{2} (s) + N_{2} H_{4} (a q) ⟶ C u (s) + N_{2} (g) + 2 H_{2} O$
So, one mole of $N_{2} H_{4}$ reduces two moles of $C u (O H)_{2}$ .

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Similar questions

Q. In the following redox reaction:

2 C u {(O H)}_{2} (s) + N_{2} H_{4} (a q) ⟶ 2 C u (s) + N_{2} + 4 H_{2} O (g)

The number of moles of

C u {(O H)}_{2}

reduced by one mole of

N_{2} H_{4}

Balance the following redox reactions in acidic solutions:

B r O_{3}^{-} + N_{2} H_{4} ⟶ {B r}^{-} + N_{2}

Q. Consider the following reaction equilibrium:

N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)

. Initially, 1 mole of

N_{2}

H_{2}

are taken in a 2 litre flask.

At equilibrium state, if the number of moles of

N_{2}

is 0.6, then the total number moles of all gases present in the flask are:

H N O_{3}

4

moles of electrons in a redox reaction. The reduced product of this reaction is :

H N O_{3}

absorbs one mole of electrons in a redox reaction. The reduced product of this reaction is :

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Action of Water on Calcium Carbide

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