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 and temperature are held constant?

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}n=

RT

pV

​

R = 0.08206 L×atm/mol×K

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

0.08206×463

1.036×48.4

​

=1.319mol

According to the reaction:

n(H2S) = \frac{2}{3}

3

2

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

3

2

​ ×1.319=0.8798mol

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

p

nRT

​ V(H

2

​ S)=

1.036

0.8798×0.08206×463

​ =32.26L

Answer: 32.26 L

Step-by-step explanation: