The product of the complete combustion of any fuel (in this case, acetylene) are indeed water and carbon dioxide.
Balancing the combustion reaction,
C2H2 +(5/2) O2 --> 2CO2 + H2O
The number of moles of C2H2 will be,
(12 g) x (1 mole/26 g) = 6/13 mole
Then, the number of moles of O2 is,
(12 g) x (1 mole/32 g) = 3/8 mole
Therefore the limiting reaction is the O2. Getting the amount of CO2 and H2O produced from balancing,
CO2 = (3/8 moles) x (2 moles CO2/ 5/2 mole O2)(44 g/ 1 mole) = 52.8 g
H2O = (3/8 moles) x (1 mole / 5/2 mole O2)(18 g / 1 mole) = 2.7 g
When finding the moles in a compound you have to know the grams. In this case, 25.2 grams are given for KMnO4. To find the moles you would divide the amount of grams by the molar mass of KMnO4. The molar mass of KmnO4 is 158.034. You you would now divide 25.2 by 158.034 which is 0.15946 moles. Depending on how many decimal places the questions asks for is dependent on you. I just went with 5 significant figures.
