Question

$HCl+NaOH\to NaCl+H_{2}O+xcal$
$H_{2}S{O}_4+2NaOH\to N{a}_{2}S{O}_4+2H_{2}O+ycal$

In the given reactions, what is the relation between $x$ and $y$ ?

• A. $x=y$
• B. $x=2y$
• C. $x=\frac{y}{2}$
• D. $x=3y$

Answer 1

The correct option is C $x=\frac{y}{2}$

The first equation is given as :

$HCl+NaOH\to NaCl+H_{2}O+xcal$ ---$\left(i\right)$

In this reaction, a strong acid reacts with a strong base, and salt formation takes place along with the release of $1H_{2}O$ molecule and $xcal$ of heat.

The second equation is given as :

$H_{2}S{O}_4+2NaOH\to N{a}_{2}S{O}_4+2H_{2}O+ycal$ ----$\left(ii\right)$

In this reaction, $2H_{2}O$ molecules are formed.

On multiplying the first equation by $2$ to get an equal number of water molecules as in equation $\left(ii\right)$, we get :

$(HCl+NaOH\to NaCl+H_{2}O+xcal)\times 2$

$\Rightarrow 2HCl+2NaOH\to 2NaCl+2H_{2}O+2xcal$ ---$\left(iii\right)$

Now, from equations $\left(ii\right)$ and $\left(iii\right)$, we get :

$y=2x$

$\Rightarrow 2x=y$

$\Rightarrow x=\frac{y}{2}$

Hence, option (C) is the correct answer.