1. Let’s compose the equation of interaction of magnesium with nitrogen:
3Mg + N2 = Mg3N2;
2.Calculate the chemical amount of nitrogen:
n (N2) = V (N2): Vm = 33.6: 22.4 = 1.5 mol;
3. make up the proportion and find the amount of magnesium spent on the reaction:
1 mol of N2 reacts with 3 mol of Mg;
with x mol Mg – 1.5 mol N2;
x = 3 * 1.5: 1 = 4.5 mol;
4. find the mass of magnesium:
m (Mg) = n (Mg) * M (Mg) = 4.5 * 24 = 108 g.
Answer: 108 g.
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