Asked by Robyn
Determine how many grams of water can be produced from excess propane C3H8 and 2.55g oxygen? MM 02 = 32 and MM H2O = 18.2.
I tried 2.5 gm O2 X 5 mole O2/32.00 gm O2 x 4 mole H2O/5 mole O2 = 1.40. This is close but not correct. The answer is 1.15. Can you help with this?
I tried 2.5 gm O2 X 5 mole O2/32.00 gm O2 x 4 mole H2O/5 mole O2 = 1.40. This is close but not correct. The answer is 1.15. Can you help with this?
Answers
Answered by
bobpursley
balance the reaction:
C3H8+O2>> 3CO2 + 4H2O
so, you get 4/5 mole water for each mole of O2.
figure moles of O2: 2.55/32
then moles of H2O: 4/5 * 2.55/32
then grams of water: moleswater*18
Yes, I have rounded off molemasses, you do it more accurately.
I get 1.15 grams.
C3H8+O2>> 3CO2 + 4H2O
so, you get 4/5 mole water for each mole of O2.
figure moles of O2: 2.55/32
then moles of H2O: 4/5 * 2.55/32
then grams of water: moleswater*18
Yes, I have rounded off molemasses, you do it more accurately.
I get 1.15 grams.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.