The Foundations of Chemistry

Thus, 33.0 grams of CO 2 is the most CO 2 that can be produced from 16.0 grams of

CH 4 and 48.0 grams of O 2. If we had based our calculation on CH 4 rather than O 2 , our

answer would be too big (44.0 grams) and wrongbecause more O 2 than we have would

be required.

Another approach to problems like Example 3-8 is to calculate the number of moles

of each reactant:

__? mol CH 4 16.0 g CH 4 1.00 mol CH 4

__? mol O 2 48.0 g O 2 1.50 mol O 2

Then we return to the balanced equation. We first calculate the required ratioof reactants

as indicated by the balanced chemical equation. We then calculate the available ratioof

reactants and compare the two:

Required Ratio Available Ratio



We see that each mole of O 2 requires exactly 0.500 mol of CH 4 to be completely used

up. We have 0.667 mol of CH 4 for each mole of O 2 , so there is more than enough CH 4

to react with the O 2 present. That means that there is insufficientO 2 to react with all of

the available CH 4. The reaction must stop when the O 2 is gone; O 2 is the limiting reac-

tant, and we must base the calculation on it. (If the available ratio of CH 4 to O 2 had been

smaller thanthe required ratio, we would have concluded that there is not enough CH 4

to react with all of the O 2 , and CH 4 would have been the limiting reactant.)

0.667 mol CH 4



1.00 mol O 2

1.00 mol CH 4



1.50 mol O 2

0.500 mol CH 4



1.00 mol O 2

1 mol CH 4



2 mol O 2

1 mol O 2



32.0 g O 2

1 mol CH 4



16.0 g CH 4

98 CHAPTER 3: Chemical Equations and Reaction Stoichiometry

Problem-Solving Tip:Choosing the Limiting Reactant

Students often wonder how to know which ratio to calculate to help find the limiting

reactant.

(1)The ratio must involve the two reactants with amounts given in the problem.

(2)It doesn’t matter which way you calculate the ratio, as long as you calculate both

required ratio and available ratio in the same order. For example, we could

calculate the required and available ratio of in the approach we just

illustrated. If you can’t decide how to solve a limiting reactant problem, as a last

resort, do the entire calculation twice—once based on each reactant amount

given. The smalleranswer is the right one.

mol O 2



mol CH 4

EXAMPLE 3-9 Limiting Reactant

What is the maximum mass of Ni(OH) 2 that could be prepared by mixing two solutions that

contain 25.9 g of NiCl 2 and 10.0 g of NaOH, respectively?

NiCl 2 
2NaOH88nNi(OH) 2 
2NaCl