Question

In the reaction ${Cr}_5+H_2{SO}_4\to {Cr}_2({SO}_4{)}_3+H_{2}O+O_2$ one mole of ${CrO}_5$ will liberate how many moles of $O_2$:

• A. $5/2$
• B. $5/4$
• C. $9/2$
• D. $7/4$

Answer 1

The correct option is D $7/4$

$2Cr{O}_5+3H_{2}S{O}_4⟶C{O}_2(S{O}_4{)}_3+3H_{2}O+\frac{7}{2}{O}_2$

As $2$ moles of $Cr{O}_5$ gives $\frac{7}{2}{O}_2$; $1$ mole of $Cr{O}_5$ gives $\frac{7}{4}{O}_2$ .