Question

In the reaction,
$Cr{O}_5+H_{2}S{O}_4\to C{r}_2(S{O}_4{)}_3+H_{2}O+O_2$
one mole of $Cr{O}_5$ will liberate how many moles of $O_2$ ?

• A. 7/4
• B. 5/4
• C. 9/2
• D. None of these

Answer 1

The correct option is A 7/4

$4Cr{O}_5+6H_{2}S{O}_4\to 2C{r}_2(S{O}_4{)}_3+7O_2+6H_{2}O$

Accorrding to the above balenced equation 4 moles of $Cr{O}_5$ gives with 7 moles of $O_2$
So 1 mole of $Cr{O}_5$ gives 7/4 moles.
Hence option A is correct.