Question

$CuO+\left(X\right)\to CuS{O}_4+H_{2}O$

Here, $\left(X\right)$ is:

• A. $CuS{O}_4$
• B. $HCl$
• C. $H_{2}S{O}_4$
• D. $HN{O}_3$

Answer 1

The correct option is C $H_{2}S{O}_4$

Acid and base react together, neutralize each other and form salt and water.

$Acid+Base\to Salt+water$

Here $CuO$ acts as a base and the salt formed is $CuS{O}_4$, along with formation of water $\left(H_{2}O\right)$.

Since the salt formed is $CuS{O}_4$, the reacting acid should be sulphate ion ($S{O}_4^{2-}$) donor, and water would not have been formed if $CuOandCuS{O}_4$ would have reacted together. Thus $H_{2}S{O}_4$ is the acid that will react with $CuO$ to give $CuS{O}_{4}and{H}_{2}O$. The reaction occurs as follows :-

$CuO+H_{2}S{O}_4\to CuS{O}_4+H_{2}O$

Hence, the correct answer is option (C).