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8.00g of CH4 is allowed to burn in the presence of 6.00g of oxgen

Resolved Question:

suppose 8.00g of CH4 is allowed to burn in the presence of 6.00g of oxgen. How much CH4,O2 and H2O remain after the reaction is complete.
Submitted: 9 years ago.
Category: Homework
Expert:  abozer replied 9 years ago.


CH4 + 2O2 → CO2 + 2H2O

(1 mole CH4 is 16 grams, 1 mole O2 is 32 grams, 1 mole H2O is 18 grams, 1 mole CO2 is 44 grams)

8.00 g of CH4 is 8/16 = 0.5 moles

0.5 moles CH4 needs 1 mole of O2 (32.00 g) to completely react. But we only have 6.00 grams of O2 so O2 will be completely reacted and some CH4 will remain at the end of the reaction.

6.00 g O2 is 6.00/32.00 = 0.1875 moles

0.1875 moles of O2 reacts with 0.1875/2 = 0.09375 moles of CH4 whic is 0.09375*16 = 1.5 grams

So at the end 8-1.5 = 6.5 grams of CH4 will be left.

No Oxygen will be left.

0.1875 moles of H2O will be formed which is 0.1875*18 = 3.375 grams

Good Luck!

Customer: replied 9 years ago.
how much CO2 in grams will be formed also from this
Expert:  abozer replied 9 years ago.
0.09375*44 = 4.125 grams of CO2 will be formed
Expert:  abozer replied 9 years ago.

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