This problem starts by giving you some information (also shown in ex. 7 in
your workbook): methane (formula CH4) burns in air (that means it reacts with
oxygen, O2) to form carbon dioxide (which has the formula CO2) and
also form water vapor (which has the formula H2O). The question is, "If 20
g of methane, CH4, was mixed with 30 g of O2 in a closed container
and ignited, what would be the limiting reagent? How much of the other one would be left
over? How much CO2 and how much H2O would be made?"
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| Methane (CH4) burns in air (reacts with O2) to form
carbon dioxide (CO2) and water vapor (H2O). If 20 g of CH4
was mixed with 30 g 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? |
|
| We will need to go through a sequence of steps to solve this problem. So,
the first thing we need is to write down the equation for the reaction and balance it. Everything
in the question is in grams, so we don't really need to write down the mole relationship,
but it helps in the next step.
We need the weight relationships, which are shown. A very useful thing at this
point, just to double check your math is to realize that the conservation of mass applies.
So, the total weight of reactants should be equal to the total weight of the products.
Checking that will serve as a double check on the math that you've done so far. Sixteen
plus 64 is 80 and 44 plus 36 is also 80, so that checks out.
|
| Balanced equation |
CH4 + |
2 O2 |
 |
CO2 + |
2 H2O |
| Mole relationship |
1 |
2 |
|
1 |
2 |
| Weight relationship |
16.0 g |
64.0 g |
|
44.0 g |
36.0 g |
|
|
Now, we need to determine the limiting reagent in the same way that you
did a few lessons ago. In this case that can be done using a rough approximation. We have
20 g of methane mixed with 30 g of oxygen. What we have available is half-again as much
oxygen as we have methane. However, if you look at the weight relationship, you can see
that it takes about four times as much oxygen. So we're going to run out of oxygen. We
just don't have enough oxygen to react with all of the methane that's available. The
limiting reagent is 30 g of oxygen.
|
(a) Determine LR by rough comparison:
Methane needs to react with 4x as much oxygen by weight.
But ony 1½ x as much is available.
Therefore oxygen is LR (and methane is excess reagent). |
|
To answer the question, "How much excess reagent would be left
over?," we will have to know how much methane (which is the excess reagent) actually
reacts. So you need to find out how much methane reacts with 30 g of oxygen. The
difference between that and the 20 g that is available will be the amount that's left
over. The calculations are shown.
|
(b) Determine methane used based on LR:
| 30 g O2 x |
16.0 g CH4
64.0 g O2 |
= 7.5 g CH4 |
used |
|
(b) Determine methane left over:
20 g available - 7.5 g used = 12.5 g left ~ 13 g left |
|
| Next we need to find out how much carbon dioxide can be made from 30 g of
oxygen and how much water can be made from 30 g of oxygen. For our purposes, let's
consider the zeroes in 20 and 30 to be significant digits. The calculations are shown. |
(c) Determine CO2 made:
| 30 g O2 x |
44.0 g CO2
64.0 g O2 |
= 21 g CO2 |
|
|
|
|
(d) Determine H2O made:
| 30 g O2 x |
36.0 g H2O
64.0 g O2 |
= 17 g H2O |
|
|
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