Question

In the following redox reaction:
$2Cu{\left(OH\right)}_2\left(s\right)+N_2{H}_4\left(aq\right)⟶2Cu\left(s\right)+N_2+4H_{2}O\left(g\right)$
The number of moles of $Cu{\left(OH\right)}_2$ reduced by one mole of $N_2{H}_4$ is :

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$2$

The balanced reaction is as follows:
$2Cu{\left(OH\right)}_2\left(s\right)+N_2{H}_4\left(aq\right)⟶2Cu\left(s\right)+N_2\left(g\right)+4H_{2}O$
So, one mole of $N_2{H}_4$ reduces two mole of $Cu(OH{)}_2$.