Question

Which volume of ethane gas, in cm3 will produce 40 cm3 of carbon dioxide gas when mixed with 140 cm3 of oxygen gas, assuming the reaction goes to completion?

2C2H6 (g) + 7O2 (g) → 4CO2 (g) + 6H2O (g)

Answer 1

201.6 cm³ is the required volume of ethane.

Given,

2C₂H₆ (g) + 7O₂ (g) → 4CO₂ (g) + 6H₂O (g)

Volume of CO₂ = 40 cm³ = 0.04 L

Volume of O₂ = 140 cm³ = 0.14 L

To Find,

Volume of C₂H₆

Solution,

We can convert volume into moles by dividing it by 22.4

Moles of CO₂ =  $\frac{0.04}{22.4}$ = 0.0018 moles

By stoichiometry -

2 moles of ethane produce 4 moles of carbon dioxide

Thus, 1-mole carbon dioxide is produced by 0.5 moles of ethane

0.0018 CO₂ → 0.0018 * 0.5 C₂H₆

Moles of ethane = 0.0018 * 0.5

Moles of ethane = 0.0009 moles

The volume of Ethane  = 0.0009 * 22.4

The volume of Ethane = 0.02016 L

The volume of Ethane = 201.6 cm³

201.6 cm³ is the required volume of ethane.

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