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5.0 g of $N_{2} (g)$ and 2.0 g of $H_{2} (g)$ are mixed to produce ${NH}_{3} (g) .$ The mass of the product obtained in this reaction is :

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5 g

6 g

4 g

3 g

answer is B.

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Detailed Solution

1 mole N₂----2 mole NH₃
0.178 mole----

$\frac{0.178 X 2}{1} = 0.358 m o l N H_{3}$
0.358X17=6.052g NH₃

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