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Chemistry Help.....again

hhh
(edited 6 years ago)
Original post by FloralPrints
hhh

moles=mass/molecular mass
1.76g/44=0.04mol CO2
0.90g/18=0.05mol H20

If we multiply 0.01 mol by 100 to get 1 mole of the hydrocarbon, we can determine the stoichiometry of the equation to be:
CxHy + 13O2 -> 4CO2 + 5H2O
(As we’ve multiplied the moles of CxHy by 100, we do the same to the products, giving 4 moles of carbon dioxide and 5 moles of water, and so by deduction we can determine the hydrocarbon to be butane)
Hope this helps! :smile:
Original post by Connorx1999
moles=mass/molecular mass
1.76g/44=0.04mol CO2
0.90g/18=0.05mol H20

If we multiply 0.01 mol by 100 to get 1 mole of the hydrocarbon, we can determine the stoichiometry of the equation to be:
CxHy + 13O2 -> 4CO2 + 5H2O
(As we’ve multiplied the moles of CxHy by 100, we do the same to the products, giving 4 moles of carbon dioxide and 5 moles of water, and so by deduction we can determine the hydrocarbon to be butane)
Hope this helps! :smile:


Thank you very much! Sorry for wasting your time as i figured out the answer right after my post but couldnt figure out how to delete it!

Thanks again and have a good night! Xxxx

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