| Here is a very similar kind of problem except in this case we work with
moles. (It is also given in ex. 8 in your workbook.) We have different starting
information but then the same questions as before. |
| Methane (CH4) burns in air (reacts with O2) to form
carbon dioxide (CO2) and water vapor (H2O). If 2.0 moles of CH4
was mixed with 3.0 moles of O2 in a closed container and ignited: a. What
would be the limiting reagent?
b. How much excess reagent would be left over?
c. How much CO2 would be made?
d. How much H2O would be made? |
|
Therefore, the equation to balance is the same one that we had before.
The mole relationships are shown by the coefficients: one mole of methane reacts with two
moles of oxygen to form one mole of carbon dioxide and two moles of water.
|
| Balanced equation |
CH4 + |
2 O2 |
 |
CO2 + |
2 H2O |
| Mole relationship |
1 |
2 |
|
1 |
2 |
|
|
| Next we need to determine the limiting reagent, and carry out the
calculations. The calculations are shown. Look through them to make sure you can follow
along. |
(a) Determine LR by rough comparison:
2.0 moles of CH4 would need 4.0 moles of O2.
But only 3.0 moles of O2 is available.
Therefore O2 is LR (and CH4 is excess reagent). |
|
|
(b) Determine methane used based on LR:
| 3.0 moles O2 x |
1 mole CH4
2 moles O2 |
= 1.5 moles CH4 |
|
(b) Determine methane left over:
2.0 moles available - 1.5 moles used = 0.5 mole left |
|
|
(c) Determine CO2 made:
| 3.0 moles O2 x |
1 mole CO2
2 moles O2 |
= 1.5 moles CO2 |
|
|
|
(d) Determine H2O made:
| 3.0 moles O2 x |
2 moles H2O
2 moles O2 |
= 3.0 moles H2O |
|
|
