When 100.g Mg3N2 reacts with 75.0 g H2O, what is the maximum theoretical yield of NH3?

Mg3N2(s) + 6H2O (l) → 3Mg(OH)2(aq) + 2NH3(g)

Answer choices:

1280 g

70.8 g

213 g

23.6 g

25.0 g

Answer : The correct option is, 23.6 g

Explanation : Given,

Mass of = 100.0 g

Mass of = 75.0 g

Molar mass of = 101 g/mol

Molar mass of = 18 g/mol

First we have to calculate the moles of and .

and,

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

From the balanced reaction we conclude that

As, 6 moles of react with 1 mole of

So, 4.17 moles of react with moles of

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of

From the reaction, we conclude that

As, 6 moles of react to give 2 moles of

So, 4.17 moles of react to give mole of

Now we have to calculate the mass of

Molar mass of = 17 g/mole

Therefore, the maximum theoretical yield of is, 23.6 grams.

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