Question #193979

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?

Expert's answer

2H_{2}S + 3O_{2} → 2SO_{2} + 2H_{2}O

V(O_{2}) = 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(O_{2}) "= \\frac{1.036 \\times 48.4}{0.08206 \\times 463}=1.319 \\; mol"

According to the reaction:

n(H_{2}S) = "\\frac{2}{3}" n(O_{2}) = "\\frac{2}{3} \\times 1.319 = 0.8798 \\;mol"

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

Answer: 32.26 L

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