$(N H_{4})_{2} C r_{2} O_{7} (s) h e a t ⟶ N_{2} (g) + 4 H_{2} O (g) + C r_{2} O_{3} (s)$

The quantity in moles of nitrogen formed in the above reaction is_______.
(Please give your answer in the form of the nearest integer)

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Solution

Gram molecular mass of $(N H_{4})_{2} C r_{2} O_{7}$ = $252$ g.
Gram molecular mass of $N_{2}$ = $28$ g.
Gram molecular mass of $C r_{2} O_{3}$ = $152$ g.

If $252$ g of $(N H_{4})_{2} C r_{2} O_{7}$ release $28$ g of $N_{2}$ = $28$ g

then, $63$ g of $(N H_{4})_{2} C r_{2} O_{7}$ will release $= \frac{28 \times 63}{252} = 7 g$ .

No. of moles of $N_{2}$ = $= \frac{7}{14} = 0.5 m o l e s \approx 1$ .

Hence, the correct answer is $1$ .

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