The Foundations of Chemistry

Solution

The balanced equation tells us that the reaction ratio is 1 mol of H 2 SO 4 to 2 mol of NaOH.

H 2 SO 4 
2NaOH88nNa 2 SO 4 
2H 2 O

1 mol 2 mol 1 mol 2 mol

From the volume and the molarity of the H 2 SO 4 solution, we can calculate the number of

moles of H 2 SO 4.

__? mol H 2 SO 4 0.0400 L H 2 SO 4 soln0.0202 mol H 2 SO 4

The number of moles of H 2 SO 4 is related to the number of moles of NaOH by the reaction

ratio, 1 mol H 2 SO 4 /2 mol NaOH:

__? mol NaOH0.0202 mol H 2 SO 4 0.0404 mol NaOH

Now we can calculate the volume of 0.505 MNaOH solution that contains 0.0404 mol of

NaOH:

__? L NaOH soln0.0404 mol NaOH0.0800 L NaOH soln

which we usually call 80.0 mL of NaOH solution.

We have worked through the problem stepwise; let us solve it in a single setup.



LH

av

2

a

S

i

O

lab

4

l

s

e

oln

 88n 

m

a

o

v

l

a

H

ila

2

b

S

l

O

e

^4 88n 

s

m

o

o

ln

lN

ne

a

e

O

de

H

d

 88n 

so

L

ln

N

n

a

e

O

ed

H

ed



__? L NaOH soln0.0400 L H 2 SO 4 soln



 0.0800 L NaOH soln or 80.0 mL NaOH soln

You should now work Exercise 78.

1.00 L NaOH soln



0.505 mol NaOH

2 mol NaOH



1 mol H 2 SO 4

0.505 mol H 2 SO 4



L H 2 SO 4 soln

1.00 L NaOH soln



0.505 mol NaOH

2 mol NaOH



1 mol H 2 SO 4

0.505 mol H 2 SO 4



L soln

The volume of H 2 SO 4 solution is
expressed as 0.0400 L rather than
40.0 mL.

112 CHAPTER 3: Chemical Equations and Reaction Stoichiometry

Key Terms

Actual yield The amount of a specified pure product actually
obtained from a given reaction. Compare with Theoretical
yield.
Chemical equation Description of a chemical reaction by plac-
ing the formulas of reactants on the left and the formulas of
products on the right of an arrow. A chemical equation must
be balanced; that is, it must have the same number of each kind
of atom on both sides.
Concentration The amount of solute per unit volume or mass
of solvent or of solution.

Dilution The process of reducing the concentration of a solute

in solution, usually simply by adding more solvent.

Limiting reactant A substance that stoichiometrically limits the

amount of product(s) that can be formed.

Molarity (M) The number of moles of solute per liter of solu-

tion.

Percent by mass 100% multiplied by the mass of a solute di-

vided by the mass of the solution in which it is contained.

Percent yield 100% times actual yield divided by theoretical

yield.

Again we see that molarity is a unit
factor. In this case,

1.00 L NaOH soln



0.505 mol NaOH