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.
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?
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