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If 54.o grams of water are mixed with magnesium nitride, how many grams of ammonia are produced

User Jimit Tank
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1 Answer

4 votes

Answer

17.02 grams

Step-by-step explanation

Given:

Mass of water = 54.0 grams

What to find:

The grams of ammonia produced.

Step-by-step solution:

The first step is to write a balanced equation for the reaction:

Mg₃N₂ + 6H₂O → 2NH₃ + 3Mg(OH)₂

From the Periodic Table:

Molar mass of water = 18.01528 g/mol

Molar mass of NH₃ = 17.031 g/mol

From the balanced equation above; 1 mol Mg₃N₂ mixed with 6 mol H₂O to produced 2 mol NH₃

In grams; 6 mol x 18.01528 g/mol = 108.09168 g H₂O produced 2 mol x 17.031 g/mol = 34.062 g NH₃

Therefore, 54.0 g H₂O will produce


\frac{54.0\text{ g }H_2O*34.062\text{ g }NH_(3)\text{ }}{108.09168\text{ g }H_2O}=17.01655484g\approx17.02\text{ grams }NH_3\text{ }

User Matcygan
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