Question

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

Answer 1

What volume of ammonia is evolved when ammonium chloride is passed through 1.48 grams of calcium hydroxide?

Solution: The equation of chemical reaction between ammonium chloride and calcium hydroxide is:

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

We will assume that the quantity of ammonium chloride is in excess to the substance amount of calcium hydroxide, and all calcium hydroxide fully reacts with ammonium chloride.

As you see, 1 mol of calcium hydroxide produces 2 moles of gaseous ammonia. According to the Avogadro's law, at the STP 1 mole of gas has a volume of 22.4 liters.

We will calculate the amount of substance of obtained ammonia:

$$n(NH_3) = 2n(Ca(OH)_2) = 2 \frac{m(Ca(OH)_2)}{M(Ca(OH)_2)},$$

where $n, m, M$ are the amount of substance, mol, mass, g, molar mass, g/mol, respectively. $M(Ca(OH)_2) = 40 + 2 \cdot 17 = 74\ \mathrm{g/mol}$;

$$n(NH_3) = 2 \cdot \frac{1.48}{74} = 0.04\ \mathrm{mol};$$

Then, volume of obtained ammonia at the STP is $V(NH_3) = n(NH_3) \cdot V_m = 0.04 \cdot 22.4 = 0.9$ liters.

Answer: 0.9 liters.