Nitrogen reacts with red-hot magnesium to form magnesium nitride,
3 Mg(s) + N₂(g) $\rightarrow$ Mg₃N₂(s)
Which reacts with water to form magnesium hydroxide and ammonia,
Mg₃N₂(s) + 6 H₂O(l) $\rightarrow$ 3 Mg(OH) ₂(aq) + 2 NH₃(aq)
How many grams of magnesium would you have to start with to prepare 15.0 grams of ammonia?

Question

Nitrogen reacts with red-hot magnesium to form magnesium nitride,
3 Mg(s) + N₂(g)

\rightarrow

Mg₃N₂(s)
Which reacts with water to form magnesium hydroxide and ammonia,
Mg₃N₂(s) + 6 H₂O(l)

\rightarrow

3 Mg(OH) ₂(aq) + 2 NH₃(aq)
How many grams of magnesium would you have to start with to prepare 15.0 grams of ammonia?

Quizplus · Accepted Answer

First, calculate the molar mass of ammonia (NH3): N (14.01 g/mol) + 3*H (3*1.008 g/mol) = 17.034 g/mol. For 15.0 g of NH3, the moles of NH3 are 15.0 g / 17.034 g/mol = 0.8805 mol. From the balanced chemical equation, 1 mol of Mg3N2 produces 2 mol of NH3, so 0.8805 mol NH3 would come from 0.44025 mol of Mg3N2. The molar mass of Mg3N2 is 3*Mg (3*24.305 g/mol) + 2*N (2*14.01 g/mol) = 100.93 g/mol. The moles of Mg needed are 3 times the moles of Mg3N2 because 3 moles of Mg produce 1 mole of Mg3N2, so 0.44025 mol of Mg3N2 corresponds to 1.32075 mol of Mg. The mass of Mg required is 1.32075 mol * 24.305 g/mol = 32.1 g, which rounds to 32.2 g, matching option D.

Nitrogen Reacts with Red-Hot Magnesium to Form Magnesium Nitride, 3 →\rightarrow→

Nitrogen Reacts with Red-Hot Magnesium to Form Magnesium Nitride,
3 $\rightarrow$