XXXXXXX
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX(^66)
(^66)
g
g
This shows that 1 mol of CH 4 reacts with 2 mol of O 2. These two quantities are chemically
equivalent, so we can construct unit factors.
Solution
CH 4 2O 2 88nCO 2 2H 2 O
1 mol 2 mol 1 mol 2 mol
? mol CH 4 24.0 g CH 4 1.50 mol CH 4
? mol O 2 1.50 mol CH 4 3.00 mol O 2
?g O 2 3.00 mol O 2 96.0 g O 2
All these steps could be combined into one setup in which we convert
g of CH 4 88n mol of CH 4 88n mol of O 2 88n g of O 2
?g O 2 24.0 g CH 4 96.0 g O 2
The same answer, 96.0 g of O 2 , is obtained by both methods.
You should now work Exercise 18.
The question posed in Example 3-4 may be reversed, as in Example 3-5.
EXAMPLE 3-5 Mass of a Reactant Required
What mass of CH 4 , in grams, is required to react with 96.0 grams of O 2?
Plan
We recall that one mole of CH 4 reacts with two moles of O 2.
Solution
__?g CH 4 96.0 g O 2 24.0 g CH 4
You should now work Exercise 22.
This is the amount of CH 4 in Example 3-4 that reacted with 96.0 grams of O 2.
EXAMPLE 3-6 Mass of a Product Formed
Most combustion reactions occur in excess O 2 , that is, more than enough O 2 to burn the sub-
stance completely. Calculate the mass of CO 2 , in grams, that can be produced by burning 6.00
mol of CH 4 in excess O 2.
16.0 g CH 4
1 mol CH 4
1 mol CH 4
2 mol O 2
1 mol O 2
32.0 g O 2
32.0 g O 2
1 mol O 2
2 mol O 2
1 mol CH 4
1 mol CH 4
16.0 g CH 4
32.0 g O 2
1 mol O 2
2 mol O 2
1 mol CH 4
1 mol CH 4
16.0 g CH 4
Here we solve the problem in three
steps and so we convert
- g CH 4 nmol CH 4
- mol CH 4 nmol O 2
- mol O 2 ng O 2
3-2 Calculations Based On Chemical Equations 95These unit factors are the reciprocals
of those used in Example 3-4.It is important to recognize that the
reaction must stop when the 6.0 mol
of CH 4 has been used up. Some O 2
will remain unreacted.