If insufficient oxygen is available, carbon monoxide can be a product of the combustion of butane: 9C4H10 + 2O2 -- 8CO + 10H2O. what mass of CO could be produced from 5.0g butane? HOW DO I DO THIS PLEASE HELP!!!!!

Question

If insufficient oxygen is available, carbon monoxide can be a product of the combustion of butane: 9C4H10 + 2O2 --> 8CO + 10H2O. what mass of CO could be produced from 5.0g butane? HOW DO I DO THIS PLEASE HELP!!!!!

anonymous · Answer

Ok my chemistry is a little rusty, but I'll try my best to explain this. 
You always start with what your given: 5.0g of butane. 
You want to end up with CO.
So you do dimensional analysis:
G of butane --> mol butane/g butane --> mol CO/ mol butane --> g CO/ mol Co

anonymous · Answer

is the answer 2.15g??

anonymous · Answer

Ezah is correct, this is a limited reagent problem, you know that got the gram of butane (C4H10) so divide that into molecular mass of which is 
Carbon 4 atoms times 12 AMU = 48
hydrogen 10 atoms times 1.007 AMU= 10 +/-
= 58grams molecular mass
58g for 1 mol of Butane/5g
so you do a mol to mol of each one, you are solving. 9 mole of C4H10 
and 8 mols of CO. from here it is easy.

anonymous · Answer

gotta know the molecular mass of CO then it is over.

anonymous · Answer

thank you!!

anonymous · Answer

does it make sense now?

anonymous · Answer

yeah it does well at least a little bit more Thank you