# 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?

Jul 10, 2018

${C}_{2} {H}_{5} O H + 2 {O}_{2} \to 2 C {O}_{2} + 3 {H}_{2} O$

mass of $C {O}_{2}$ = 191 g

#### Explanation:

Part 1: Balancing equation:

${C}_{2} {H}_{5} O H + {O}_{2} \to C {O}_{2} + {H}_{2} O$

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 $C {O}_{2}$ on the right.

${C}_{2} {H}_{5} O H + {O}_{2} \to \textcolor{red}{2} C {O}_{2} + {H}_{2} O$

Updating the atom counts on the right:
L-- Atom--R
2 -- C -- $\textcolor{red}{\cancel{1}} 2$
6 -- H -- 2
3 -- O -- $\textcolor{red}{\cancel{3}} 5$

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 ${H}_{2} O$ on the right.

${C}_{2} {H}_{5} O H + {O}_{2} \to 2 C {O}_{2} + \textcolor{red}{3} {H}_{2} O$

Updating the atom counts on the right:
L-- Atom--R
2 -- C -- $\cancel{1} 2$
6 -- H -- $\textcolor{red}{\cancel{2}} 6$
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:

${C}_{2} {H}_{5} O H + \textcolor{red}{2} {O}_{2} \to 2 C {O}_{2} + 3 {H}_{2} O$

Updating the atom counts on the right:
L-- Atom--R
2 -- C -- $\cancel{1} 2$
6 -- H -- $\cancel{2} 6$
$5 \textcolor{red}{\cancel{3}}$-- O -- $\cancel{3} 5$

Part 2: Find mass $C {O}_{2}$ from mass ethanol, ${C}_{2} {H}_{5} O H$

The plan is to perform the following 3 steps calculations starting from $\text{mass } {C}_{2} {H}_{5} O H$ to $\text{mass } C {O}_{2}$

$\text{mass " C_2H_5OH ->" moles " C_2H_5OH -> " moles " CO_2 -> "mass } C {O}_{2}$

• 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

• ${C}_{2} {H}_{5} O H = 2 \left(12\right) + 5 \left(1\right) + 16 + 1 = 46 \frac{g}{\text{mol}}$

• $C {O}_{2} = 12 + 2 \left(16\right) = 44 \frac{g}{\text{mol}}$

Calculations can be set up this way. Notice how the units cancel each other out leaving $\text{g } C {O}_{2}$:
$\text{mass of "CO_2=100" g } {C}_{2} {H}_{5} O H$$\cdot \left(1 \text{ mol "C_2H_5OH)/(46" g } {C}_{2} {H}_{5} O H\right)$*(2" mol "CO_2)/(1" mol "C_2H_5OH)*(44" g "CO_2)/(1" mol "CO_2)=191" g " CO_2