Firstly, make sure the equation is correctly balanced – which it is here; comparing right sight with left: Left=Right i.e.: 2 carbon atoms, 6 hydrogen a, 7 oxygen a
Next make it clear what is your unknown, so e.g. write x=CO2 [grams]
Now, as these are the atoms which are rearranging themselves in the reaction, we need to calculate molar mass (M) of the 2 compounds in the equation – ethanol (as we know its weight) and carbon dioxide (as it is our x).
Molar mass is calculated by adding molar masses of elements that constitute a compound. The molar mass of each element can be found in periodic table (usually in the top right corner). We need molar masses of carbon, oxygen and hydrogen and these are, accordingly: 12, 16 and 1 grams/mol.
Thus M of ethanol is: 212 + 61 + 1* 16 = 46 [g/mol]
M of carbon dioxide: 112 + 216 = 44 [g/mol]
Moreover, as the equation shows, 1 mol of ethanol gives us 2 moles of carbon dioxide.
Now, to calculate mass of CO2 formed we have to use cross product of a proportion, which says that:
From 46 g/mol of ethanol we get 88 g/mol of carbon dioxide (44*2 as we get 2 moles)
So, if we use 100 g of ethanol, how many g of carbon dioxide (x) would we get?
It can be written in the following way:
46 g/mol ethanol – 88 g/mol carbon dioxide
100 g ethanol – x g carbon dioxide
Using cross product to get x:
(100g*88) /46 = 191.3 [g] (Note that the g/mol eliminate, thus leaving us with just grams).
We thus found x and the final answer is that from 100 g of ethanol that is burnt we get 191.3 grams of carbon dioxide.