Help on an AP Chemistry Problem- need help soon?

Question

I have stumbled alot and have gotten no where PLEASE HELP! need soon!

A 9.780 g. gaseous mixture contains ethane (C2H6) and propane (C3H8). Complete combustion to form carbon dioxide and water requires 1.120 mol of oxygen. Calculate the mass percent of ethane in the original mixture.

thank you thank you thank you, I am sure to someone this is a fairly simple problem.. I really appreciate it I have spent all summer on all of my work this is one of four I really just can't get but I have a test coming up!

Answer 1

I actually didn't find this problem to be that simple! ;-)

First, determine the balanced equations for complete combustion of ethane and propane...

For Ethane:

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

and for Propane:

1 C3H8 + 5 O2 --> 3 CO2 + 4 H2O

If 100% of the gas is ethane, then 1.12 moles O2 would be needed to combust 0.32 moles ethane.

This was determined as follows...

1.12 moles O2 / 7 moles O2 = 16% O2

2 moles ethane x 0.16 = 0.32 moles ethane.

0.32 moles x 30 g/mole = 9.6g ethane

If 100% of the gas is propane, then 1.12 moles O2 would be needed to combust 0.22 moles propane.

This was determined as follows...

1.12 moles O2 / 5 moles O2 = 0.224 % O2

1 mole propane x 0.224 = 0.224 moles propane.

0.224 moles x 44 g/mol = 9.85 g propane

Therefore, if 100% of the gas was ethane, then 9.6 g of gas would be combusted with 1.12 moles of O2 and if 100% of the gas is propane, then 9.85 g of gas would be combusted with 1.12 moles of O2.

70% propane (9.85g x 0.7 = 6.90g) + 30% ethane  (9.60g x 0.3 = 2.88g) = 9.78 g

Therefore, the answer to the question is 30% ethane...

Hope this helps!!!