JEE MAIN - Chemistry (2022 - 29th June Morning Shift - No. 15)
The number of terminal oxygen atoms present in the product B obtained from the following reaction is _____________.
FeCr2O4 + Na2CO3 + O2 $$\to$$ A + Fe2O3 + CO2
A + H+ $$\to$$ B + H2O + Na+
Answer
6
Explanation
$\mathrm{FeCr}_{2} \mathrm{O}_{4}+\mathrm{Na}_{2} \mathrm{CO}_{3}+\mathrm{O}_{2} \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}+\mathrm{CO}_{2}$ $+\mathrm{Na}_{2} \mathrm{CrO}_{4}$
(A)
$$ \mathrm{Na}_{2} \mathrm{CrO}_{4}+\mathrm{H}^{+} \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{-2}+\mathrm{H}_{2} \mathrm{O}+\overset{+}{\mathrm{Na}} $$
_29th_June_Morning_Shift_en_15_1.png)
$$ \mathrm{Na}_{2} \mathrm{CrO}_{4}+\mathrm{H}^{+} \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{-2}+\mathrm{H}_{2} \mathrm{O}+\overset{+}{\mathrm{Na}} $$
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