Question

Ammonium sulfate reacts with potassium hydroxide solution to produce ammonia gas, potassium sulfate, and water. Calculate the volume of ammonia gas, measured at 23oC and 64 kPa, that could be produced from 264.0 g of ammonium sulfate and 280.0 g of potassium hydroxide

Answer 1

The balanced equation of the reaction between the ammonium sulfate with potassium sulfate:

(NH₄)₂SO₄ + 2KOH $\rightarrow$ 2NH₃$\uparrow$ +K₂SO₄ + 2H₂O.

As one can see, 1 mol of ammonium sulfate reacts with 2 mol of potassium hydroxide:

$n((NH_4)_2SO_4 = \frac{n(KOH)}{2}$ .

The number of the moles of ammonium sulfate in 264.0 g can be calculated using its molar mass (132.14 g/mol):

$n((NH_4)_2SO_4 = \frac{264.0}{132.14} = 2.0$ mol.

In the same way, the number of the moles of potassium hydroxide:

$n(KOH) = \frac{280}{56.1} = 5.0$ mol.

The potassium hydroxide is in excess, because: 2/1<5/2. Assuming that all ammonium sulfate react, the number of the moles of ammonia gas is:

$n(NH_3) = 2·n((NH_4)_2SO_4) = 2·2.0 =4.0$ mol.

Making use of the ideal gas law, we can calculate the volume of the ammonia gas:

$V = \frac{nRT}{p} = \frac{4.0·8.314·(23+273.15)}{64·10^3} = 0.154$ m³, or 154 L.

Answer: 154 L of ammonia gas, measured at 23°C and 64 kPa, could be produced from 264.0 g of ammonium sulfate and 280.0 g of potassium hydroxide, assuming that all the ammonium sulfate react.