How many grams of calcium nitride is formed when 1.12 liters of nitrogen reacts with 14 grams of calcium? What substance and in what quantity is taken in excess?
V (N) = 1.12 l.
m (Ca) = 14 g.
Determine the mass of calcium nitride. We write down the solution.
First, we compose the reaction equation.
3Са + 2 N = Ca3N2.
We find the amount of nitrogen and calcium.
n = V / Vm.
n (N) = 1.12 / 22.4 = 0.05 mol.
M (Ca) = 40 g / mol.
n (Ca) = 14/40 = 0.35 mol.
Determine how many moles of nitrogen should be.
X = 0.35 * 2: 3 = 0.23 mol.
This means that nitrogen is in short supply, therefore, we count by nitrogen.
0.05 mol N – x mol Ca3N2
2 mol N – 1 mol Ca3N2
X = 0.05 * 1: 2 = 0.025 mol Ca3N2
Find the mass of calcium nitride.
M (Ca3N2) = 3 * 40 + 2 * 14 = 120 + 28 = 148 g / mol.
m = 0.025 * 148 = 3.7 g.
Answer: m (Ca3N2) = 3.7 g.
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