Calculate the volume of air containing 21 oxygen by volume required for complete combustion of ethylene
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Answer:
2C2H2+5O2⟶4CO2+2H2O
From stoichiometric calculations:
2 C2H2 moles will react with 5 molecules of oxygen
1 C2H5⟶5/2 moles of O2
22.4 l C2H2⟶ 5/2×22.4 l of O2
1 volume of air -> 0.2 (20/100) volume of oxygen
Now the only thing I don’t know is how to convert the volume into liters so that I can find the correct answer.
Explanation:
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