Question

What volume of Ammonia is evolved when Ammonium chloride is passed through 1.48 grams of CalciumHydroxide

Answer 1

Task:

What volume of Ammonia is evolved when Ammonium chloride is passed through 1.48 grams of Calcium Hydroxide

Solution:

The chemical equation of the reaction is

$$2 \mathrm{NH_4Cl} + \mathrm{Ca(OH)_2} = \mathrm{CaCl_2} + 2 \mathrm{NH_3} + 2 \mathrm{H_2O}$$

The number of moles of $\mathrm{Ca(OH)_2}$ is

$$n(\mathrm{mol}) = \frac{m(g)}{MW(g/mol)}$$ $$MW(\mathrm{Ca(OH)_2}) = AW(\mathrm{Ca}) + 2 \cdot AW(\mathrm{O}) + 2 \cdot AW(\mathrm{H}) = 40 + 2 \cdot 16 + 2 \cdot 1 = 74\ \mathrm{g/mol}$$ $$n(\mathrm{Ca(OH)_2}) = \frac{1.48}{74} = 0.02\ \mathrm{mol}$$

According to the chemical equation the number of moles of $\mathrm{NH_3}$ is

$$n(\mathrm{NH_3}) = 2 \cdot n(\mathrm{Ca(OH)_2}) = 2 \cdot 0.02 = 0.04\ \mathrm{mol}$$

The volume of $\mathrm{NH_3}$ at STP is

$$V(L) = n(\mathrm{mol}) \cdot V_0(L)$$

$V_0(L) = 22.4\ \mathrm{L}$ – the volume of 1 mol of gas at STP (p = 1 atm, T = 273 K)

$$V(L) = 0.04 \cdot 22.4 = 0.896\ \mathrm{L}$$

Answer: $V(L) = 0.896\ \mathrm{L}$