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?

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

32.26 L

Explanation:

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Explanation:

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