How do you balance $C_{2} H_{5} O H (l) + O_{2} (g) \to C O_{2} (g) + H_{2} O (g)$ $C_2H_5OH(l) + O_2(g) -> CO_2(g) + H_2O(g)$ and determine the mass of $C O_{2}$ $CO_2$ produced from the combustion of 100.0 g ethanol?

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How do you balance $C_{2} H_{5} O H (l) + O_{2} (g) \to C O_{2} (g) + H_{2} O (g)$ $C_2H_5OH(l) + O_2(g) -> CO_2(g) + H_2O(g)$ and determine the mass of $C O_{2}$ $CO_2$ produced from the combustion of 100.0 g ethanol?

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Answer 1 · 2018-07-10T00:22:27

$C_{2} H_{5} O H + 2 O_{2} \to 2 C O_{2} + 3 H_{2} O$ $C_2H_5OH +2O_2 -> 2CO_2 + 3H_2O$

mass of $C O_{2}$ $CO_2$ = 191 g

Explanation:

Part 1: Balancing equation:

$C_{2} H_{5} O H + O_{2} \to C O_{2} + H_{2} O$ $C_2H_5OH + O_2 -> CO_2 + H_2O$

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 $C O_{2}$ $CO_2$ on the right.

$C_{2} H_{5} O H + O_{2} \to 2 C O_{2} + H_{2} O$ $C_2H_5OH + O_2 ->color(red) 2CO_2 + H_2O$

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 $"g "CO_2$ :
$"mass of "CO_2=100" g " C_2H_5OH$ $*(1" mol "C_2H_5OH)/(46" g " C_2H_5OH)$ $*(2" mol "CO_2)/(1" mol "C_2H_5OH)*(44" g "CO_2)/(1" mol "CO_2)=191" g " CO_2$

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How do you balance $C_{2} H_{5} O H (l) + O_{2} (g) \to C O_{2} (g) + H_{2} O (g)$ $C_2H_5OH(l) + O_2(g) -> CO_2(g) + H_2O(g)$ and determine the mass of $C O_{2}$ $CO_2$ produced from the combustion of 100.0 g ethanol?

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Explanation:

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How do you balance C_2H_5OH(l) + O_2(g) -> CO_2(g) + H_2O(g)C2H5OH(l)+O2(g)→CO2(g)+H2O(g)C_2H_5OH(l) + O_2(g) -> CO_2(g) + H_2O(g) and determine the mass of CO_2CO2CO_2 produced from the combustion of 100.0 g ethanol?

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How do you balance $C_{2} H_{5} O H (l) + O_{2} (g) \to C O_{2} (g) + H_{2} O (g)$ $C_2H_5OH(l) + O_2(g) -> CO_2(g) + H_2O(g)$ and determine the mass of $C O_{2}$ $CO_2$ produced from the combustion of 100.0 g ethanol?