# 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. One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.