How do you balance $C_2H_5OH(l) + O_2(g) -> CO_2(g) + H_2O(g)$ and determine the mass of $CO_2$ produced from the combustion of 100.0 g ethanol?

Part 1: Balancing equation:

$C_2H_5OH + O_2 -> CO_2 + H_2O$

We'll do that by counting the number of atoms on the left and right hand side of the equation and get them to be the same.

L-- Atom--R
2 -- C -- 1
6 -- H -- 2
3 -- O -- 3

Looks like C and H are not balanced. Since there are 2 C on the left and only 1 C on the right, we'll add 2 in front of $CO_2$ on the right.

$C_2H_5OH + O_2 ->color(red) 2CO_2 + H_2O$

Updating the atom counts on the right:
L-- Atom--R
2 -- C -- $color(red)cancel(1)2$
6 -- H -- 2
3 -- O -- $color(red)cancel(3)5$

Now that C is balanced, we'll move on to H. There are 6 H on the left and only 2 H on the right, we'll add 3 in front of $H_2O$ on the right.

$C_2H_5OH + O_2 -> 2CO_2 + color(red)3H_2O$

Updating the atom counts on the right:
L-- Atom--R
2 -- C -- $cancel(1)2$
6 -- H -- $color(red)cancel(2)6$
3 -- O -- $cancel(3)5$

All that's left now is to balance O. We have 3 O on the left and 5 O on the right. If we add a 2 in front of $O_2$ on the left, we'll balance O:

$C_2H_5OH + color(red)2O_2 -> 2CO_2 + 3H_2O$

Updating the atom counts on the right:
L-- Atom--R
2 -- C -- $cancel(1)2$
6 -- H -- $cancel(2)6$
$5color(red)cancel(3)$-- O -- $cancel(3)5$

Part 2: Find mass $CO_2$ from mass ethanol, $C_2H_5OH$

The plan is to perform the following 3 steps calculations starting from $"mass " C_2H_5OH$ to $"mass "CO_2$

$"mass " C_2H_5OH ->" moles " C_2H_5OH -> " moles " CO_2 -> "mass "CO_2 $

• Converting mole to mole will require the stoichiometric coefficient (numbers in front of balanced equation).
• Converting from mass to mole of the same substance will require its molar mass.

In that case, we'll need to find the molar mass of

• $C_2H_5OH=2(12)+5(1)+16+1= 46 g/"mol"$

• $CO_2=12+2(16)= 44 g/"mol"$

Calculations can be set up this way. Notice how the units cancel each other out leaving $"g "CO_2$:
$"mass of "CO_2=100" g " C_2H_5OH$$*(1" mol "C_2H_5OH)/(46" g " C_2H_5OH)$$*(2" mol "CO_2)/(1" mol "C_2H_5OH)*(44" g "CO_2)/(1" mol "CO_2)=191" g " CO_2$