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?

1 Answer
Jul 10, 2018

Answer:

#C_2H_5OH +2O_2 -> 2CO_2 + 3H_2O#

mass of #CO_2# = 191 g

Explanation:

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#