Answer:
-125.4
Explanation:
Target equation is 4C(s) + 5H2(g) = C4H10
These are the data equations for enthalpy of combustion
- C(s) + O2(g) =O2(g) -393.5 kJ/mol * 4
- H2(g) + ½O2(g) =H20(l) = 285.8 kJ/mol * 5
- 2CO2(g) + 3H2O(l) = 13/2O2 (g) + C4H10 - 2877.1 reverse
To get target equation multiply data equation 1 by 4; multiply equation 2 by 5; and reverse equation 3, so...
Calculate 4(-393.5) + 5(-285.8) + 2877.6 and you should get the answer.
Answer:
oh coooool
I have one Id it is on ace with maybe 5800 points
maybe this picture will help in something
Answer:
186 Liters at STP conditions
Explanation:
1 mole of any gas at STP conditions occupies 22.4 Liters.
Therefore, 8.32 moles O₂(g) = 8.32 moles x 22.4Liters/mole = 186 Liters (3 sig.figs.)