Question

In the reaction given below, $1$ mol of $Cr{O}_5$ will liberate how many moles of $O_2$?

$Cr{O}_5+H_{2}S{O}_4\to C{r}_2(S{O}_4{)}_3+H_{2}O+O_2$

• A. $\frac{5}{2}$
• B. $\frac{5}{4}$
• C. $\frac{9}{2}$
• D. None of the above

Answer 1

The correct option is D None of the above

The balanced reaction is as follows:
$2Cr{O}_5+3H_{2}S{O}_4\to C{r}_2(S{O}_4{)}_3+3H_{2}O+\frac{7}{2}{O}_2$
$2$ $3$ $1$ $3$ $\frac{7}{2}$

$2$ mol of $Cr{O}_5$ liberates $\frac{7}{2}$ mol of $O_2$.
$\therefore$ $1$ mol $Cr{O}_5$ liberates $\frac{7}{4}$ mol of $O_2$.