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

The quantity of $(N H_{4})_{2} C r_{2} O_{7}$ in moles if 163 gm of $(N H_{4})_{2} C r_{2} O_{7}$ is heated as per the given equation is_______.

(Please give your answer in the form of nearest integer)

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Solution

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

$N o . o f m o l e s = \frac{W e i g h t o f (N H_{4})_{2} C r_{2} O_{7}}{G r a m m o l e c u l a r m a s s}$

$= \frac{163}{252} = 0.65 m o l e \approx 1$ .

Hence, the correct answer is $1.$

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(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)

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

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

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[N = 14, H = 1, O = 16, C r = 52]

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