Final answer:
To produce 87 g of magnesium hydroxide from magnesium nitride, approximately 50 g of magnesium nitride is needed according to stoichiometry calculations based on the balanced chemical reaction provided.
Step-by-step explanation:
The task is to determine the amount of magnesium nitride needed to produce 87 g of magnesium hydroxide, given the reaction between magnesium nitride and water which yields magnesium hydroxide and ammonia gas:
Mg3N2 (s) + 6H2O(l) → 3Mg(OH)2 (s) + 2NH3(g)
To solve this problem, we can use stoichiometry. First, we find the molar mass of Mg(OH)2 is approximately 58.32 g/mol. Then we calculate the number of moles of Mg(OH)2:
87 g Mg(OH)2 ÷ 58.32 g/mol = 1.492 moles Mg(OH)2
Next, looking at the balanced reaction, for every 3 moles of Mg(OH)2 produced, 1 mole of Mg3N2 is consumed. We calculate the moles of Mg3N2 needed:
1.492 moles Mg(OH)2 × (1 mole Mg3N2 / 3 moles Mg(OH)2) = 0.497 moles Mg3N2
The molar mass of Mg3N2 is roughly 100.93 g/mol, so we can determine the mass of Mg3N2 needed:
0.497 moles Mg3N2 × 100.93 g/mol = 50.162 g Mg3N2
The closest answer to this calculated mass is 50 g of magnesium nitride, which is option B.