Question

A 5.00 g mixture of methane (CH4) and ethane (C2H6) is combusted in oxygen gas to produce carbon dioxide and water. If 14.09 g of CO2 is produced, how many grams of methane was in the original 5.00 g mixture ?

Answer 1

Solution :-

Combustion equations are

CH4 + 2O2 ---- > CO2 + 2H2O

2C2H6 + 7O2 ---- > 4CO2 + 6H2O

Mixture = 5.00 g

Methane gives 1 mol CO2 and 2 mol ethane gives 4 moles of CO2

Total mass of CO2 = 14.09 g

So lets assume mass of methane = x gram

Then mass of ethane = 5.00 – x g

So lets make the set up

(mass of CH4 * mm of CO2/mm CH4)+(5.00-x)*(4* mm CO2 / 2*mm ethane) = 14.09 g

(x*44.01/16.04) + ((5.00-x)*(4*44.01 g / 2*30.069 g)) = 14.09

2.74376 x + ((5.00 –x)*2.92726) = 14.09 g

Solve for the x we get

2.977 g = x

So the mass of CH4 (methane) in the original sample = 2.977 g we can round it to 2.98 g