Magnesium interacts with 33.6 liters of nitrogen. Determine the mass of metal spent

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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