Chapter 5: Stoichiometry Part 4 SABIS Grade 11 (Level M) Chemistry ~ SABIS ® Curriculum Chemistry Online Tuition

Chapter 5: Stoichiometry

5.4 Solving Problems

5.4.1 Sulfuric Acid as an Oxidizing Agent

One of the important uses of sulfuric acid is that of an oxidizing agent. For example

when heated, it will even dissolve carbon. The unbalanced equation is:

C + H2SO4 ⇌ CO2 + H2O + SO2

5.4.2 Sulfuric Acid in Aqueous Solution

A second major use of sulfuric acid is in reactions with bases. In laboratory use it is diluted to a much lower concentration and can be used as a standard acid. A typical problem would be the titration of a base solution of unknown concentration using a sulfuric acid solution of known concentration.

Molarity of solutions: Revision

The unit of concentration of solutions is molarity. A molar solution contains 1 mol/dm3. The relation between number of moles in a given volume of a certain concentration is given by:

n (in mols) = V (in dm3) × c (in mol/dm3)

A hint for solving problems on molarity

Frequently, a problem requires two or more steps to solve. Consider the problem:

Find moles of acid to react with V cm3 of c M base
How many moles of HCl react with 250 cm3 of 0.80 M KOH?

The first step is to find how much base there is in the given base solution.
Amount of base given = 250 cm3 × 0.80 mol/ 1000 cm3 = 0.20 mol

The second step is to use a balanced equation to calculate the acid needed:

HCl + KOH → KCl + H2O

Since the acid and base react in the ratio of 1 mole to 1 mole, 0.20 mol of acid is needed.

Whenever a two-step problem gives specified volume with a certain concentration, this is equivalent to giving the number of moles. Therefore the problem may be simplified by taking the product of these two terms to obtain the number of moles. Here are two examples.

1. How many moles of H2SO4 react with 400 cm3 of 2.0 M NaOH?

“400 cm3 of 2.0 M NaOH” can be simplified to (0.400 dm3 × 2.0 M) = 0.80 mol

So the problem is simplified to:

How many moles of H2SO4 react with 0.80 mol NaOH?

2. What volume of 0.50M HCl react with 28.4 ml of 1.0 M NaOH?

Here the problem can be simplified twice:

a) What volume of 0.50 M HCl can be simplified to how many moles. Once we calculate that, we can use n = V × c to calculate the volume.

b) 28.4 ml of 1.00 M NaOH can be simplified to (28.4 ml × 1.00 M) = 28.4 m-mol.

(Notice that the symbol m means “milli” or 10–3, and it does not interfere with the other unit, which is “mol”. In other words, m-mol means 10–3 mol).

So the problem can be simplified to:
How many moles of HCl react with 28.4 m-mol NaOH?

Example

What is the concentration of a sodium hydroxide solution if 25.45 ml of the NaOH solution just reacts with 18.55 ml of 0.125 M H2SO4 (to produce a neutral solution)?

Given: 25.45 ml NaOH react with 18.55 ml of 0.125 M H2SO4.
RTF: [NaOH]

Notice that this problem can be simplified to “How many moles of NaOH react with (18.55 × 0.125) = 2.32 m-mol of H2SO4?

Solution

NaOH + H2SO4→ Na2SO4 + H2O

Step 1: Balance the equation, determine reaction ratio:

2NaOH + H2SO4→ Na2SO4 + 2H2O
2 moles + 1 mole

Step 2:

Moles of H2SO4 used = 18.55 ml × 0.125 mol/l = 2.32 m-mol

Moles of NaOH needed = n = 2.32 m-mol H2SO4 × $\frac{2 m o l N a O H}{1 m o l H_{2} S O_{4}}$ = 4.64 m-mol

Step 3:

To find the concentration: c(NaOH) = n/V = 4.64 m-mol / 25.45 ml = 0.182 M

What volume of 0.5 M sulfuric acid is needed to neutralize 40 ml of 0.5 M KOH?

Give your answer, without the unit, as an integer rounded to the nearest units.

Volume of sulfuric acid = ..... cm3

5.4.3 Sulfuric Acid Reacts with some Metals to Give Hydrogen

In the reaction:

C + H2SO4 ⇌ CO2 + H2O + SO2

Sulfuric acid acts as an oxidizing agent. It oxidizes carbon and sulfur is reduced (check the oxidation number). When it reacts with, say, zinc, sulfuric acid also acts as an oxidizing agent, but in this case hydrogen is reduced:

Zn + H2SO4 → H2 + ZnSO4

1 mole + 1 mole 1 mole

Example

Calculate the volume of hydrogen obtained over water on a day when the atmospheric pressure is 767 mm Hg and the temperature is 27°C, when 6.54 g of zinc are added to 250 ml of 0.500 M H2SO4. Vapor pressure of water at 27°C is 27 mm Hg.

Given: Mass of Zn = 6.54 g; Quantity of H2SO4 = 250 ml of 0.500M.

Pwater = 27 mm Hg. Room temp & pressure = 27°C and 767 mm Hg

RTF: Volume of hydrogen produced.

Solution

To know the volume of H2 we have to find the number of moles H2 produced.

Step 1: Since we have two quantities of reactants given, we need to know which one finishes completely:

Zn + H2SO4 → H2 + ZnSO4
1 mole 1 mole 1 mole

Given quantity of Zn = 6.54 g × $\frac{1 m o l}{65.4 g}$ = 0.100 mol

(Remember that you are actually given the number of moles of H2SO4.)

Given quantity of H2SO4 = 250 ml × $\frac{0.500 m o l}{1000 m o l}$ = 0.125 mol

Reaction ratio of Zn / H2SO4 = 1/1

Given ratio of Zn / H2SO4 = 1 / 1.25

So Zn is the smaller (relative) quantity, and it reacts completely. It determines the amount of hydrogen produced.

Step 2: Determine the quantity of H2 produced:

Quantity of H2 produced= 0.100 mol Zn × $\frac{1 m o l H_{2}}{1 m o l Z n}$ = 0.100 mol H2

Step 3: Determine the volume of H2 produced:

In PV = nRT, we have

n = 0.100 mol H2;

R = 0.0821 $\frac{d m^{3} a t m}{K m o l}$ ;

P = 767 – 27 = 750 mm Hg = 740/760 atm

T = 27°C = 300 K

V = ?

V = nRT/P = 0.1 × 0.0821 × 300 × 760/740 = 2.53 dm3

What volume of hydrogen gas would be produced at room temperature and pressure when 3.27 g of Zn react completely with excess sulfuric acid?

Given Molar mass in g.mol−1: Zn = 64

Give your answer to two significant figures.

Volume of H2 = .... dm3

Chapter 5: Stoichiometry Part 4 SABIS Grade 11 (Level M) Chemistry