How much amount of $N H_{3}$ is produced when 28 g of nitrogen completely react with 6 g of hydrogen?

32.5 g

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6 g

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28 g

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34 g

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Solution

The correct option is D 34 g
According to the law of conservation of mass, the total mass of the products is equal to that of reactants, in a chemical reaction.

$3 H_{2} (g) + N_{2} (g) \to 2 {NH}_{3} (g) 6 g 28 g X g$

The mass of $N H_{3}$ produced is equal to the sum of mass of hydrogen and nitrogen that reacted to give $N H_{3}$ .

Hence, the mass of $N H_{3}$ = X g = 28 + 6 = 34 g.

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How much amount of NH3 is produced when 28 g of nitrogen completely react with 6 g of hydrogen?

How much amount of $N H_{3}$ is produced when 28 g of nitrogen completely react with 6 g of hydrogen?