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antoniya [11.8K]
3 years ago
5

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

Chemistry
1 answer:
Temka [501]3 years ago
4 0

Answer : The correct option is, 23.6 g

Explanation : Given,

Mass of Mg_3N_2 = 100.0 g

Mass of H_2O = 75.0 g

Molar mass of Mg_3N_2 = 101 g/mol

Molar mass of H_2O = 18 g/mol

First we have to calculate the moles of Mg_3N_2 and H_2O.

\text{Moles of }Mg_3N_2=\frac{\text{Given mass }Mg_3N_2}{\text{Molar mass }Mg_3N_2}

\text{Moles of }Mg_3N_2=\frac{100.0g}{101g/mol}=0.990mol

and,

\text{Moles of }H_2O=\frac{\text{Given mass }H_2O}{\text{Molar mass }H_2O}

\text{Moles of }H_2O=\frac{75.0g}{18g/mol}=4.17mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

Mg_3N_2(s)+6H_2O(l)\rightarrow 3Mg(OH)_2(aq)+2NH_3(g)

From the balanced reaction we conclude that

As, 6 moles of H_2O react with 1 mole of Mg_3N_2

So, 4.17 moles of H_2O react with \frac{4.17}{6}=0.695 moles of Mg_3N_2

From this we conclude that, Mg_3N_2 is an excess reagent because the given moles are greater than the required moles and H_2O is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NH_3

From the reaction, we conclude that

As, 6 moles of H_2O react to give 2 moles of NH_3

So, 4.17 moles of H_2O react to give \frac{2}{6}\times 4.17=1.39 mole of NH_3

Now we have to calculate the mass of NH_3

\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3

Molar mass of NH_3 = 17 g/mole

\text{ Mass of }NH_3=(1.39moles)\times (17g/mole)=23.6g

Therefore, the maximum theoretical yield of NH_3 is, 23.6 grams.

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