I need help with this question. Determine the mass of water vapor that forms when a lighter burn 1.00 g of butane, C 4 H 10?

Question

jfraser · Answer

what's the balanced equation for the combustion of butane?

(a combustion reaction always takes the fuel, combines it with molecular oxygen, and always produces carbon dioxide and water

anonymous · Answer

for a complete reaction of butane it requires oxygen.
there fore, C4H10 + 13 O = 4Co2 + 5H2O will come.... find find out number of moles will come to butane... you can get it by wt/mol. wt. formula and find the number of moles required as oxygen... and then balance them with the right hand side products in the mole basis itself.. you will get the answer after converting the resulted moles of h2o into wt again.....

jfraser · Answer

@harishk molecular oxygen is always $O_2$, not just $O$.

anonymous · Answer

Thanks, i figure out and my balance quation is 2C4HY10+5O2 = 10H2O+ 8C
I did ( 1,00g C4HY10 *1mol C4HY10 * 10mol H2O * 18g/mol H2O) / (58g C4H10 *2mol C4H10 * 1 mol H2O). And it gave me about ~ 1,55g of H2O.

anonymous · Answer

sorrry ... my bad