When 20 g of ammonium chloride interacted with Ca hydroxide, a gas was formed

When 20 g of ammonium chloride interacted with Ca hydroxide, a gas was formed, which was burned in O2. Determine the number of molecules formed in a simple substance.

During the first reaction:
2NH4Cl + Ca (OH) 2 = CaCl2 + 2H2O + 2NH3 ↑.
From the reaction equation:
ν (NH4Cl) = ν (NH3).
Knowing that ν = m / M, we have:
m (NH4Cl) / M (NH4Cl) = V (NH3) / Vm (NH3).
Vm (NH3) = 22.4 L / mol.
Let’s calculate the molar mass of ammonia chloride:
M (NH4Cl) = 14 + 1 * 4 + 35.5 = 53.5 g / mol.
Let’s determine the amount of ammonia that will be released during the reaction:
V (NH3) = m (NH4Cl) * Vm (NH3) / M (NH4Cl).
Substitute the numerical values:
V (NH3) = 20 * 22.4 / 53.5 = 8.37 liters.

Combustion of ammonia in oxygen:
4NH3 + 3O2 = 2N2 + 6H2O.
In the course of the reaction, nitrogen will be a simple substance at the exit, we will find how much it turned out:
ν (NH3) / 4 = ν (N2) / 2.
V (NH3) / (4 * Vm (NH3)) = ν (N2) / 2.
Let us express the amount of nitrogen substance:
ν (N2) = V (NH3) * 2 / (4 * Vm (NH3)).
Substitute the numbers and get:
ν (N2) = V (NH3) * 2 / (4 * Vm (NH3)) = 8.37 * 2 / (4 * 22.4) = 0.187 mol.
The amount of a substance, through the number of molecules, is determined by the expression:
ν = N / Na, where Na is Avogradro’s constant, Na = 6.02 * 10 ^ 23 1 / mol.
Expressing the number of molecules from this expression and substituting the numbers, we get:
N = ν * Na = 0.187 * 6.02 * 10 ^ 23 = 1.125 * 10 ^ 23.
Answer: 1.125 * 10 ^ 23 nitrogen molecules will be released.