Ammonia can be produced in a two-step process. First, nitrogen reacts with red-hot magnesium to form magnesium nitride. Then magnesium nitride reacts with water to form magnesium hydroxide and ammonia.
A. Write each balanced chemical reaction.
B. How many grams of magnesium would we have to start with to prepare 15.0 grams of ammonia?
A. 3Mg(s) + N2(g) --> Mg3N2(s)
Mg3N2(s) + 6H2O --> 3Mg(OH)2(aq) + 2NH3(aq)
B. 15.0gNH3 x (1moleNH3 / 17.03056gNH3) x (1moleMg3N2 / 2molesNH3) x (100.92848gMg3N2 / 1moleMg3N2) = 54.0gMg3N2
54.0gMg3N2 x (1moleMg3N2 / 100.92848gMg3N2) x (3moleMg / 1moleMg3N2) x (24.3050gMg / 1moleMg) = 39.0gMg
Is this right?
Yes, your calculations are correct.
In part A, you correctly wrote the balanced chemical reactions for the two steps in the process:
Step 1: 3Mg(s) + N2(g) --> Mg3N2(s)
Step 2: Mg3N2(s) + 6H2O --> 3Mg(OH)2(aq) + 2NH3(aq)
In part B, you used the given mass of ammonia (15.0g) and performed a series of conversion factors to calculate the mass of magnesium needed. Your calculations are correct:
15.0g NH3 x (1 mol NH3 / 17.03056g NH3) x (1 mol Mg3N2 / 2 mol NH3) x (100.92848g Mg3N2 / 1 mol Mg3N2) = 54.0g Mg3N2
Then, you used another set of conversion factors to calculate the mass of magnesium:
54.0g Mg3N2 x (1 mol Mg3N2 / 100.92848g Mg3N2) x (3 mol Mg / 1 mol Mg3N2) x (24.3050g Mg / 1 mol Mg) = 39.0g Mg
Therefore, to produce 15.0 grams of ammonia, you would need to start with approximately 39.0 grams of magnesium.