2CH3OH + 3O2 ==> 2CO2 + 4H2O
dHrxn in kJ = (n*dHf products( - (n*dHf reactants)
dHrxn x (49g/2*molar mass CH3OH) = ?kJ.
What is the enthalpy change (in kJ) for burning 49.0 g of methyl alcohol?
I tried this question and got the wrong answer of 1047.74.
A gas is compressed from 73.7 L to 1.9 L at 0.70 atmospheres? In addition a total of 4400 J of heat are gained from warmer surroundings. What is the change in internal energy in kJ?
I have a general idea of how to do this, but I tried and got -2851.8, which isn't right.
Any help is appreciated!
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