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Stoichiometric Calculations

For per gram ...

Question

For per gram of reactant, the maximum quantity of $N_{2}$ gas is produced in which of the following thermal decomposition reactions?

[Given: Atomic wt. $C r = 52 u, B a = 137 u$ ].

B a (N_{3})_{2} (s) \to B a (C) + 3 N_{2} (g)

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

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2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g)

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2 N H_{4} N O_{3} (s) \to 2 N_{2} (g) + 4 H_{2} O (g) + O_{2} (g)

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Solution

The correct option is C $2 N H_{3} (g) \to N_{2} (g) + 3 H_{2} (g)$
(A)
Molar mass of $B a (N_{3})_{2} (s) = 221$ g/mol.

1 mole of $B a (N_{3})_{2} (s)$ will give 3 moles of $N_{2}$

$\frac{1 g}{221 g / m o l}$ moles of $B a (N_{3})_{2} (s)$ will give $3 \times \frac{1}{221} = 0.014$ moles of $N_{2}$
(B)
Molar mass of $(N H_{4})_{2} C r_{2} O_{7} = 252$ g/mol.

1 mole of $(N H_{4})_{2} C r_{2} O_{7}$ will give 1 mole of $N_{2}$

$\frac{1 g}{252 g / m o l}$ moles of $(N H_{4})_{2} C r_{2} O_{7}$ will give $1 \times \frac{1}{252} = 0.0039$ moles of $N_{2}$
(C)
Molar mass of $N H_{3} = 17$ g/mol.

2 mole of $N H_{3}$ will give 1 mole of $N_{2}$

$\frac{1 g}{17 g / m o l}$ moles of $N H_{3}$ will give $\frac{1}{2 \times 17} = 0.0297$ moles of $N_{2}$
(D)
Molar mass of $N H_{4} N O_{3} = 80$ g/mol.

1 mole of $N H_{4} N O_{3}$ will give 1 mole of $N_{2}$

$\frac{1 g}{80 g / m o l}$ moles of $N H_{4} N O_{3}$ will give $1 \times \frac{1}{80} = 0.0125$ moles of $N_{2}$

Hence, the correct option is $C$

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Similar questions

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