The Foundations of Chemistry

94 CHAPTER 3: Chemical Equations and Reaction Stoichiometry

EXAMPLE 3-3 Mass of a Reactant Required

What mass of oxygen is required to react completely with 1.20 mol of CH 4?

Plan

The balanced equation

CH 4 
 2O 2 88nCO 2 
 2H 2 O

1 mol 2 mol 1 mol 2 mol

16.0 g 2(32.0 g) 44.0 g 2(18.0 g)

gives the relationships among moles and grams of reactants and products.

mol CH 4 88n mol O 2 88n g O 2

Solution

__?g O 2 1.20 mol CH 4 76.8 g O 2

EXAMPLE 3-4 Mass of a Reactant Required

What mass of oxygen is required to react completely with 24.0 g of CH 4?

Plan

Recall the balanced equation in Example 3-3.

CH 4 
 2O 2 88nCO 2 
 2H 2 O

1 mol 2 mol 1 mol 2 mol

16.0 g 2(32.0) g 44.0 g 2(18.0) g

32.0 g O 2



1 mol O 2

2 mol O 2



1 mol CH 4

Problem-Solving Tip:Use the Mole Ratio in Calculations with

Balanced Chemical Equations

The most important way of interpretating the balanced chemical equation is in terms of

moles. We use the coefficients to get the mole ratio of any two substances we want to

relate. Then we apply it as

moles of moles of mole ratio

desired  substance  from balanced

substance given chemical equation

It is important to include the substance formulas as part of the units; this can help

us decide how to set up the unit factors. Notice that in Example 3-2, we want to cancel

the term mol CH 4 , so we know that mol CH 4 must be in the denominator of the mole

ratio by which we multiply; we want mol O 2 as the answer, so mol O 2 must appear

in the numerator of the mole ratio. In other words, do not just write 

m

m

o

o

l

l

; write

, giving the formulas of the two substances involved.

mol of something



mol of something else

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