Question

Hydrogen sulfide gas is combusted with oxygen gas to produce sulfur dioxide gas and water vapour. If there is 48.4 L of oxygen available, what volume of hydrogen sulfide gas may be combusted if the pressure is held constant at 105 kPa and the temperature is held constant at 190°C?

Answer 1

2H₂S + 3O₂ → 2SO₂ + 2H₂O

V(O₂) = 48.4 L

p = 105 kPa = 1.036 atm

T = 190 + 273 = 463 K

Ideal gas law:

pV = nRT

$n = \frac{pV}{RT}$

R = 0.08206 L×atm/mol×K

n(O₂) $= \frac{1.036 \times 48.4}{0.08206 \times 463}=1.319 \; mol$

According to the reaction:

n(H₂S) = $\frac{2}{3}$ n(O₂) = $\frac{2}{3} \times 1.319 = 0.8798 \;mol$

$V = \frac{nRT}{p} \\ V(H_2S) = \frac{0.8798 \times 0.08206 \times 463}{1.036}=32.26 \;L$

Answer: 32.26 L