Under $S T P$ , $22.4$ liters of $N_{2} (g)$ are made to combine completely with a sufficient quantity of $H_{2} (g)$ . How much volume in litres of $N H_{3} (g)$ will be produced?

11.2

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22.4

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44.8

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67.2

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89.6

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Solution

The correct option is A $44.8$
$N_{2} + 3 H_{2} \to 2 N H_{3}$
The volume of the gas is directly proportional to the number of moles of gas. Hence, the stoichiometric mole ratio is equal to the stochiometric volume ratio.
As can be seen from balanced chemical equation, 1 mole $N_{2}$ reacts to give 2 moles of ammonia. Hence, 22.4 $L$ of nitrogen will react to give $2 \times 22.4 = 44.8 L$ of ammonia at $S T P$ .

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67.2

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

Calculate the volume of ammonia produced. What is the other substances, if any, that remains in the resultant mixture?

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Under STP, 22.4 liters of N2(g) are made to combine completely with a sufficient quantity of H2(g). How much volume in litres of NH3(g) will be produced?

Under $S T P$ , $22.4$ liters of $N_{2} (g)$ are made to combine completely with a sufficient quantity of $H_{2} (g)$ . How much volume in litres of $N H_{3} (g)$ will be produced?