How many grams of calcium nitride is formed when 1.12 liters of nitrogen reacts with

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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