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
mass of
Explanation:
Part 1: Balancing equation:
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
Updating the atom counts on the right:
L-- Atom--R
2 -- C --
6 -- H -- 2
3 -- O --
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
Updating the atom counts on the right:
L-- Atom--R
2 -- C --
6 -- H --
3 -- O --
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
Updating the atom counts on the right:
L-- Atom--R
2 -- C --
6 -- H --
Part 2: Find mass
The plan is to perform the following 3 steps calculations starting from
- 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