The combustion of 1 mol carbon, C(s), in the presence of oxygen gas, O2(g), to produce 1 mol of carbon dioxide gas, CO2(g), releases 393.5 kJ of . This is given by:
C(s)+O2(g)→CO2(g)
Calculate the standard enthalpy change upon the formation of 25.5 g of CO2(g) from C(s) and O2(g).
So you know you get 393.5 kJ for 1 mol (44 g) CO2. How much do you get for only 25.5 g CO2?
You can do this a couple of ways. One is to set a ratio/proportion like this.
(393.5 kJ/44 g CO2) = (x kJ/25.5 g CO2) and solve for X.
Or you can it the logical way like this:
393.5 kJ x (25.5/44) = ?. The logic goes like this.
393.5 kJ x a factor to correct for it being 25.5 and not 44 grams. So you KNOW factor must be either 25.5/44 or 44/25.5.Which to choose. Easy.
If you get 393.5 kJ for 44 grams, doesn't it seem logical that you get LESS for fewer grams. Sure it does. So you KNOW you MUST place the small number on top and the large number on the bottom. If you do it the other way; the large number on top and the small number on bottom you will get MORE.