# Chemical Calculations

## Quick Notes

**Mole:**The number of atoms in exactly 12 g of carbon-12Number of moles = n

**Avogadro's number:**The number of atoms/molecules in one mole of a substance = 6.02 x 1023**Molar mass:**The mass of one mole of a substance (element or compound)Molar mass = Mr

Mr = addition of the mass of every element in the compound

Unit = g/moles

**Law of definite proportions:**A compound always contains exactly the same proportion of elements by mass.**Percentage yield:**Measure the effectiveness of experiment**Empirical formula:**simplest form of ratio**Molar volume:**The volume occupied by 1 mole of any gas at r.t.p.**Limiting reactant:**A reactant is completely consumed in a chemical reaction**Excess reactant:**The reactants left behind in a chemical reaction

### Relative Atomic Mass

**Comparing Atomic Masses with the Carbon Atom**

To compare to a carbon atom, a carbon-12 atom is used.

The mass of the isotope is 12 times greater than hydrogen atom so of carbon-12 atoms is equivalent to the mass of one hydrogen atom.

**Relative Atomic Mass:**the average mass of one atom of the element (averaging isotopes) when compared with mass of a carbon-12 atom.The Relative Atomic Masses are already stated on the periodic table above each chemical formula.

### Relative Molecular Mass and Relative Formula Mass

Using Ar, we calculate Relative Masses of molecules and ionic compounds

**Relative Molecular Mass**

Molecules containes atoms joined together, e.g. Cl

_{2}Average mass (molecular mass) of Cl

_{2}= add relative masses of both atoms.

**Relative Molecular Mass:**the average mass of one molecule of substance (averaging isotopes) when compared with mass of a carbon-12 atom.

same as relative molecular mass but for ions only

total Ar of all atoms in formula of ionic compound

**Eg. Relative formula mass of MgSO**_{4}**?**

Mr = 24 + 32 + 4(16) = 120

### Percentage Composition

**Eg. Determine which oxides of iron of Fe**_{2}**O**_{3}** or Fe**_{3}**O**_{4}** has more iron.**

**Solution**

Mr(Fe_{2}O_{3}) = 2(56) + 3(16) = 160

Percentage of Fe in Fe_{2}O_{3}

=

x 100 %

= (2 x 56)/160 x 100%

= 70%

Mr(Fe_{3}O_{4}) = 3(56) + 4(16) = 232

Percentage of Fe in Fe_{2}O_{3} = x 100 %

= x 100%

= 72%

Therefore, Fe_{3}O_{4} has more iron composition than that of** **Fe_{2}O_{3}.

**Calculating the Mass of an Element in a Compound**

Use the example of Fe_{2}O_{3} in the example above.

The percentage mass of iron in iron(III) oxide is 70%.

Therefore to calculate mass of iron in a 200g compound of iron(III) oxide is (0.7 x 200)g = 140g

**Eg. Determine the mass of iron in 200g of Fe**_{2}**O**_{3}**.**

Mr(Fe_{2}O_{3})= 2(56) + 3(16) = 160

Mass of Fe in Fe_{2}O_{3}

=

x 200g

= (2 x 56)/160 x 200g

= 140g

**Calculating the Mass of Water in a Compound**

Compound with water mass is ‘hydrated’ and has H_{2}O in their formula.

**Eg. Calculate water mass in 12.5g hydrated copper sulfate, CuSO**_{4}**.5H**_{2}**O**

Mass of 5H_{2}O in CuSO_{4}.5H_{2}O

=

x mass of sample

= (5 x 18)/250 x 12.5g

= 4.5g

### Mole

**Counting Particles**

Unit for particles = mole

Symbol = mol

1 mol = 6 x 10

^{23}atoms

Moles of Particles - Calculating the Number of Moles

**Eg 1: How many molecules in 6 x 10**^{24}** molecules of water, H**_{2}**O?**

= 5 mol

**Eg 2: Calculate the number of molecules in 0.25 mole of CO**_{2}**. Hence, how many atoms are present?**

Number of particles = 0.25 mol x 6 x 10^{23}

= 1.5 x 10^{23} molecules

Number of atoms = total number of atoms in CO_{2} x number of particles

= 3 x 1.5 x 10^{23}

= 4.5 x 10^{23 }atoms

Molar mass

**Molar mass**– the mass of one mole of any substances

**For substances consisting of atoms**It is the Ar of the element in grams. Eg. Ar(C) = 12, molar mass = 12g

**For substances consisting of molecules**It is the Ar of the substance in grams. Eg. Ar(H

_{2}O) = 18, molar mass = 18g

**For substances consisting of ions**It is the Ar of substance in grams. Eg. Ar(NaCl)= 58.5, molar mass= 58.5g

**Eg. Find the mass of 0.4 mol of iron atom.**

n = m / Mr

m = n x Mr

m = 0.4 x 56 = 22.4 g

**Eg. Argon Fluorohydride gas, HArF, first known noble gas compound, has molar mass of 60g. Find the number of moles Argon atom in 6.66g of HArF.**

n (HArF) = 6.66/60

= 0.111 mol

n (Ar) = 0.111 mol x 1 Ar atom in HArF

= 0.111 mol

### Different Kinds of Chemical Formulae

**Molecular Formula**– shows the actual formula and kinds of atoms present, e.g. C_{2}H_{6}**Empirical Formula**– shows the simplest whole number ratio of the atoms present, e.g. C_{2}H_{6}, ratio 1:3, therefore C_{1}H_{3}, simply CH_{3}**Structural Formula**– shows how atoms are joined in the molecule. It can be represented by ball-and-stick model or diagrammatically.

Calculating the Empirical Formula of a Compound

**Find the empirical formula of an oxide of magnesium consisting of 0.32g of oxygen and 0.96g of magnesium.**

Step 1: find the number of moles of the 2 elements.

n(Mg) =0.96 / 24

= 0.04 mol

n(O) = 0.32 / 16

= 0.02 mol

Step 2: Divide the moles by the smallest number.

Mg = 0.04 / 0.02

= 2

O = 0.02 / 0.02

= 1

Therefore, the empirical formula is Mg_{2}O

Calculating the Empirical Formula from Percentage Composition

An oxide of sulphur consists of 40% sulphur and 60% oxygen.

Take the total 100% to be 100g.

Step 1: find the number of moles of the 2 elements.

n(S) = 40 / 32

= 1.25 mol

n(O) = 60 / 16

= 3.75 mol

Step 2: Divide the moles by the smallest number.

S = 1.25 / 1.25

= 1

O = 3.75 / 1.25

= 3

Therefore, the empirical formula is SO_{3}

Changing From Empirical formula to Molecular Formula

Find the molecular formula of propene, CH_{2}, having molecular mass of 42.

Molecular formula will be C_{n}H_{2n}

Relative molecular mass = 12n(from carbon Ar) + 2n(2 x hydrogen Ar) = 14n

14 n = 42

n = 42 / 14 = 3

Therefore --> C_{3}H_{6}

### Molar Volume of Gases

The Avogadro’s Law:

Equal volume of gases at same temperature and volume contain equal number of particles or molecules.

Molar Volume of Gas:

volume occupied by one mole of gas

All gases at room temperature and pressure (r.t.p.) = 24dm

^{3}1dm

^{3}= 1000cm^{3}

**Formulae:**

**Eg. What is the number of moles of 240cm**^{3}** of Cl**_{2}** at r.t.p.?**

= 0.01 mol

Molar Volume and Molar Mass

Gases have same volume but not necessarily same mass

Example: Hydrogen -> 2g, Carbon Dioxide -> 44g

**Eg. Find the volume of 7g of N**_{2 }**at r.t.p.**

Step 1: Find the number of moles from the mass of nitrogen

n = 7 / 28

= 0.25 mol

Step 2: Find the volume of nitrogen, now with formula of gas

0.25 mol = volume of gas / 24

Volume of gas = 0.25 mol x 24

= 6 dm^{3} (or 6000cm^{3})

### Concentration of Solutions

Concentration of solution tells the number of solute in a volume of solution

Calculating the Amount of Solute

Moles of solute (n) = Concentration (mol/dm^{3} ) x Volume of solution (dm^{3})

**Eg. What is the mass of solute in 600cm**^{3}** of 1.5 NaOH solution?**

Volume of solution in dm3 = 0.60 dm^{3}

n = 1.5 x 0.60

= 0.9 mol

Number of moles of NaOH = m / Mr

0.9 = m / 40

m = 0.9 x 40

= 36g

### Calculations using Chemical Equations

Constructing Chemical Equations

**Eg. 1: Reaction Between Hydrogen and Oxygen**

Word Equation: Oxygen + Hydrogen --> Water

To write the chemical equation, we use symbols of atoms/molecules:

O_{2} + H_{2} --> H_{2}O

Balance the equation!

O_{2} + 2H_{2} --> 2H_{2}O

Calculations from Equations

**Reacting Masses**

In every equation, each atom is rational to each other.

Suppose we want to find moles of X atoms that reacted to form 0.25 mole of Y atoms.

We always put the atom we want to find as numerator and the denominator being the atom we know.

**Eg. X + 2Z --> 2Y**

1. Find the ratio first:

2. Then multiply the ratio by no. of moles of Y to find the reacting mole of X.

1/2 x 0.25 = 0.125 mole

Therefore 0.125 mole of X reacted with 0.25 mole of Y.

3. To find the reacting mass of X, e.g. Y is given as 35g, we just multiply the mole by the mass of Y as they are always in ratio:

0.125 x 35 = 4.375 g

**Reacting Masses and Volumes**

First, find the ratio of moles and multiply the mole of the gas volume you want to find with the volume of gas at room temperature (24dm^{3})

**Example**

MgCl_{2} is formed by reacting Mg and HCl according to equation:

Mg (s) + 2HCl (aq) --> MgCl_{2} (s) + H_{2} (g)

Find the amount of hydrogen gas, in cm3, formed when 14.6g of HCl is reacted.

m(HCl) = 14.6/36.5

=0.4 mol

Multiply ratio by mole of HCl = 1/2 x 0.4 = 0.2 mol

Multiply mole by molar volume of gas at r.t.p. = 0.2 x 24 dm^{3}= 4.8 dm^{3}

1dm^{3} = 1000cm^{3}

Therefore, 4.8dm^{3} x 1000 = 4800 cm^{3}

4800cm^{3} of gas is formed

## Example Questions

**1 mole of CO**_{2}** is equal to**

- 6.02 x 10^{23} molecules

- 2 moles of O

- 1 mole of C

- 44.0095 grams

- 22.4 L at r.t.p.

- 31.9988 grams of O

- 12.0107 grams of C

**When 1.00 g of germanite was treated in this way, the germanium present was completely converted into 0.177 g of a chloride containing 33.9% by mass of germanium.**

**i. Determine the empirical formula of the chloride.**

**ii. Write down the valency of germanium in the chloride.**

**iii. Calculate the percentage of germanium in germanite**

**Solution**

ai. PbS + 4H_{2}O_{2} ---> PbSO_{4} + 4H_{2}O

aii. no. of moles of PbS = Mass / Mr

= 0.25 / (207 + 32)

= 0.001046

1 mole of PbS reacts with 4 moles of H_{2}O_{2}

0.001046 moles of PbS react with 0.001046 / 1 x 4 = 0.004184 moles of H_{2}O_{2}

Volume of H_{2}O_{2 }required = no. of moles / concentration

= 0.004184 / 0.100

= 0.0418 dm^{3}

bi.

## MCQ Questions

**1. One mole of each of the following compounds is burnt in excess oxygen. Which compound will produce three moles of carbon dioxide and three moles of steam only?**

a. C_{3}H_{8}

b. C_{3}H_{7}OH

c. C_{3}H_{7}CO_{2}H

d. CH_{3}CO_{2}CH_{3}

**2. 20 cm**^{3}** of carbon monoxide are reacted with 10 cm**^{3}** of oxygen. The equation for the reaction is shown:**

**2CO + O**_{2}** --> 2CO**_{2}

**Which volume of carbon dioxide will be produced?**

a. 10 cm^{3}

b. 20 cm^{3}

c. 30 cm^{3}

d. 40 cm^{3}

**3. What has a mass equal to that of one mole of water?**

a. 24 dm^{3 }of water

b. one mole of steam

c. one molecule of water

d. two moles of hydrogen molecules and one mole of hydrogen molecules

**4. An 8 g sample of oxygen contains the same number of atoms as 16 g of element X. What is the relative atomic mass, Ar of X?**

a. 4

b. 8

c. 16

d. 32

**5. Which quantity is the same for one mole of ethanol and one mole of ethane?**

a. mass

b. number of atoms

c. number of molecules

d. volume at r.t.p.

**6. Which ion is present in the highest concentration in a 2 mol/dm**^{3}** aqueous solution of sodium sulphate?**

a. hydrogen ions

b, hydroxide ions

c. sodium ion

d. sulphate ion

**7. Magnesium reacts with hydrochloric acid. Which solution would give the fastest initial rate of reaction?**

a. 40 g of HCl in 1000 cm^{3} of water

b. 20 g of HCl in 1000 cm^{3 }of water

c. 10 g of HCl in 100 cm^{3 }of water

d. 4 g of HCl in 50 cm^{3 }of water

**8. One mole of hydrated copper(II) sulphate, CuSO**_{4}**.5H**_{2}**O is dissolved in water. How many moles of ions does the solution contain?**

a. 1

b. 2

c. 6

d. 7

**9. What is the ratio of the volume of 2 g of hydrogen to the volume of 16 g of methane, both volumes at r.t.p.?**

a. 1 to 1

b. 1 to 2

c. 1 to 8

d. 2 to 1

**10. Which of the following contains the same number of molecules as 9 g of water?**

a. 2 g of hydrogen gas

b. 14 g of nitrogen gas

c. 32 g of oxygen gas

d. 44 g of carbon dioxide gas

**11. Calcium reacts with water as shown.**

**Ca (s) + 2H**_{2}**O (l) --> Ca(OH)**_{2}** (aq) + H**_{2}** (g)**

**What is the total mass of the solution that remains when 40g of calcium reacts with 100 g of water?**

a. 58 g

b. 74 g

c. 138 g

d. 140 g

**12. 20 cm**^{3}** of oxygen are reacted with 20 cm**^{3}** of carbon monoxide. What are the volumes of the gases remaining, at the original temperature and pressure?**

**13. What is the mass of oxygen contained in 72 g of pure water?**

a. 16 g

b. 32 g

c. 64 g

d. 70 g

**14. A volume of ethane, C**_{2}**H**_{6}**, at r.t.p. has a mass of 20 g. What is the mass of an equal volume of propene, C**_{3}**H**_{6}** at r.t.p.?**

a. 20 g

b. 21 g

c. 28 g

d. 42 g

**15. Sodium reacts with water according to the equation below.**

**2Na + 2H**_{2}**O --> 2NaOH + H**_{2}

**Which volume of hydrogen is produced at r.t.p. when 0.2 mol of sodium reacts?**

a. 1.2 dm^{3}

b. 2.4 dm^{3}

c. 4.8 dm^{3}

d. 9.6 dm^{3}

**16. What is the mass of aluminium in 204g of aluminium oxide, Al**_{2}**O**_{3}**?**

a. 26g

b. 27g

c. 54g

d. 108g

**17. The equation for the burning of hydrogen is: 2H**_{2}** (g) + O**_{2}** (g) ---> 2H**_{2}**O (g)**

**One mole of hydrogen gas is made to react with one mole of oxygen gas. What will be present after the reaction?**

a. 1 mol of steam only

b. 1 mol of steam + 0.5 mol of oxygen gas

c. 1 mol of steam + 1 mol of hydrogen gas

d. 2 mol of steam + 0.5 mol of oxygen gas

**18. The compound SO**_{2}**Cl**_{2}** reacts with water according to the equation:**

**SO**_{2}**Cl**_{2 }**+ 2**_{H2}**O ---> H**_{2}**SO**_{4}** + 2HCl**

**How many moles of sodium hydroxide will neutralise the solution produced by one mole of SO**_{2}**Cl**_{2 }**and excess water?**

a. 1

b. 2

c. 3

d. 4

**19. How many atoms are there in (NH**_{4}**)SO**_{4}**?**

a. 15

b. 14

c. 10

d. 7

**20. 3Cu + 8HNO**_{3}** --> XCu(NO**_{3}**)**_{2}** + 2NO + YH**_{2}**O**

**What are the appropriate values of X and Y for the balanced reaction equation above?**

a. X = 3, Y = 8

b. X = 4, Y = 3

c. X = 3, Y = 4

d. X = 4, Y = 8

**21. What is the percentage by mass of calcium in calcium carbonate?**

a. 4%

b. 12%

c. 40%

d. 44%

**22. Which of the following compounds contain the same percentage by mass of nitrogen as ammonium cyanate, NH**_{4}**CNO?**

a. NH_{4}NO_{3}

b. NH_{4}Cl_{2}

c. N_{2}H_{2}

d. (NH_{2})_{2}CO

**23. A hydrocarbon contains 20% of carbon and 80% of hydrogen. What is its molecular formula if it has a relative molecular mass of 30?**

a. CH_{3}

b. CH_{4}

c. C_{2}H_{6}

d. C_{2}H_{4}

**24. 0.1 mol of metal X (Ar = 27) was burned in oxygen to give an oxide with mass of 5.1g. What is the formula of the metal oxide if it has a relative molecular mass of 30?**

a. XO

b. XO_{2}

c. X_{2}O_{3}

d. X_{2}O

**25. 100cm**^{3}** of gaseous hydrogen contains n molecules. How many molecules are there in 100cm**^{3}** of gaseous methane (CH**_{4}**) under the same temperature and pressure?**

a. n

b. n/5

c. 5n/2

d. 2n/5

**26. Which of the following consists of the greatest number of atoms?**

a. 4g of H_{2}

b. 8g of O_{2}

c. 71g of Cl_{2}

d. 72dm^{3} of argon gas at rtp

**27. Which of the following consists of the most number of molecules?**

a. 32g of O_{2}

b. 18g of water

c. 28g of N_{2}

d. 4g of H_{2}

**28. Which of the following formulae for ionic compounds is not correct?**

a. Mg(OH)_{2}

b. KF

c. Al_{2}O_{3}

d. Ca_{2}Cl

**29. What is the maximum mass of chromium, Cr, that can be extracted from 76g of chromium(III) oxide?**

a. 38g

b. 48g

c. 52g

d. 152g

**30. How many oxygen atoms are present in 0.2 moles of N**_{2}**O**_{5}**?**

a. 6.02 x 10^{23}

b. 4.05 x 10^{23}

c. 3.01 x 10^{22}

d. 6.02 x 10^{21}

### MCQ Answers

1. d

2. b

3. b

4. d

5. c (Avogadro's law)

6. c

7. c

8. b

9. a

10. b

11. c (no. of moles of Ca = 40/40 = 1; no. of moles of water = 100/18. So Ca is the limiting reactant and H_{2}O was present in excess. Mass of hydrogen gas liberated/lost = 2g. Mass of solution = 140 - 2 = 138 g)

12. c

13. c

14. c

15. b

16. d

17. b

18. d

19. a

20. c

21. c

22. d

23. c

24. d

25. a

26. a

27. d

28. d

29. c

30. a

## Structured Question Worked Solutions

**1. When iron is heated in a steam of dry chlorine, it produces a chloride that contains 34.5% by mass of iron.**

**a. calculate the empirical formula of this chloride**

**b. the relative molecular mass of this chloride (Mr) is 325.**

**i. what is the molecular formula of this chloride?**

**ii. hence construct an equation, including state symbols, for the reaction of iron with chlorine.**

**Solution**

1a. molar ratio of Fe : Cl = 34.5/56 : (100 - 34.5)/35.5

= 0.616 : 1.845

= 1 : 3

Hence empirical formula is FeCl_{3}

1bi. let the molecular formula be (FeCl_{3})_{n}

n x (56 + 3 x 35.5) = 325

--> n = 2

Hence molecular formula is Fe_{2}Cl_{6}

1bii. 2Fe (s) + 3Cl_{2} (g) --> Fe_{2}Cl_{6} (s)

**2. CFCs are compounds that contain only carbon, chlorine and fluorine. They are atmospheric pollutants and destroy ozone in the upper atmosphere.**

**a. 'CFC11' has the following composition by mass**

**C: 8.7%**

**F: 13.8%**

**Cl: 77.5%**

**Calculate the empirical formula of CFC11**

**b. 'CFC12' has the molecular formula CF**_{2}**C**_{l2}**. It can be made by the reaction of hydrogen fluoride, HF, with tetrachloromethane, CCl**_{4}

**CCl**_{4}** + 2HF --> CCl**_{2}**F**_{2}** + 2HCl**

**What is the maximum mass of CFC12 that can be made from 10.0 g of hydrogen fluoride?**

**Solution**

2a.

Empirical formula: CFCl_{3}

2b. Mr of HF = 20

Mr of CF_{2}C_{l2 }= 121

2 x 20 g of HF give 121 g of CF_{2}C_{l2}

Therefore, 10 g of HF give (121/40) x 10 = 30.25 g of CF_{2}C_{l2}

**3. Potassium superoxide, KO**_{2}** is an ionic solid. It can be used in spacecraft to supply oxygen according to the following equation.**

**4KO**_{2 }**(s) + 2H**_{2}**O (l) --> 4KOH (s) + 3O**_{2}** (g)**

**The potassium hydroxide formed removes carbon dioxide.**

**a. show that 1.0 g of potassium superoxide will supply about 0.25 dm**^{3 }**of oxygen at room temperature and pressure.**

**bi. name the compound formed when carbon dioxide reacts with solid potassium hydroxide**

**bii. give the equation for the formation of this compound**

**c. Supplies of oxygen in hospitals are stored in cylinders.**

**i. state one other use for oxygen**

**ii. describe briefly how oxygen is obtained from air**

**Solution**

3a. Mr of KO_{2} = 71

4 x 71 = 284 g of KO_{2}_{ }give 3 x 24 = 72 dm^{3} of O_{2} at r.t.p.

hence, 1.0 g of KO_{2} give (72/284) x 1 = 0.25 dm^{3 }of O_{2} at r.t.p

3bi. potassium carbonate

3bii. CO_{2 }(g) + 2KOH (s) --> K_{2}CO3 (s) + H_{2}O (l)

3ci. together with acetylene in welding

3cii. oxygen is obtained by fractional distillation of liquid air

**4. Many cars are fitted with air-bags which inflate in an accident. Air-bags contain the solid sodium azide, NaN**_{3}**, which decomposes rapidly to form sodium and nitrogen. The nitrogen fills the air-bag.**

**a. construct the equation, including state symbols, for the decomposition of sodium azide.**

**b. in a crash, an air-bag fills with 72 dm**^{3}** of nitrogen at room temperature and pressure. What mass of sodium azide is needed to provide the nitrogen?**

**c. Sodium azide, NaN**_{3 }**reacts with dilute hydrochloric acid to give sodium chloride and a compound A.**

**Compound A contains 2.33% hydrogen and 97.7% nitrogen by mass.**

**i. what is the empirical formula of compound A?**

**ii. construct the equation for the reaction between sodium azide and dilute hydrochloric acid.**

**Solution**

4a. 2NaN_{3} (s) --> 2Na (s) + 3N_{2} (g)

4b. no. of moles of nitrogen = 72/24 = 3

no. of moles of NaN_{3} required = 2

mass of NaN_{3} = 2 x (23 + 3 x 14) = 130 g

4ci.

Empirical formula: N_{3}H

4cii. NaN_{3 }+ HCl --> NaCl + N_{3}H

**5. Lead white, a white pigment used in old paintings, contains lead(II) carbonate. It darkens when exposed to air containing traces of hydrogen sulphide, due to the formation of black lead(II) sulphide, PbS. The white colour can be restored by treating the painting with aqueous hydrogen peroxide which converts lead(II) sulphide into lead(II) sulphate and water.**

**ai. Write the chemical equation of the reaction between lead(II) sulphide and hydrogen peoxide.**

**aii. Calculate the volume of 0.100 mol/dm**^{3}** hydrogen peroxide required to react with 0.25 g of lead(II) sulphide.**

**b. The element germanium (Ge) was once an important component of transistors. The flow chart below shows how germanium can be made from its ore germanite.**

--> Empirical formula = GeCl_{4}

bii. valency = 4

biii. Mass of germanium in germanium chloride = 33.9 / 100 x 0.177

= 0.0600 g

% of germanium in germanite = 0.0600 / 1.00 x 100%

= 6.00% (3 sf)

**6. Calculate the number of moles of the following:**

**a. 62.1g of BaCl**_{2}

**b. H**_{2}**SO**_{4}** in 55cm**^{3}** of 0.25 mol/dm**^{3}** H**_{2}**SO**_{4}** solution**

**c. 17dm**^{3}** of nitrogen gas N**_{2}

**Solutions**

6a. Mr of BaCl_{2} = 208

no. of mol of BaCl_{2} = 62.1/208 = 0.2985 mol --> 0.3mol (2 dp)

6b. no. of mol of H_{2}SO_{4 }= concentration_{ }x volume = 0.25 x 55/1000 = 0.01375 --> 0.01 mol (2 dp)

6c. Mr of N_{2} gas = 28

no. of mol of N_{2} gas = 17/28 = 0.6071 mol --> 0.61 mol (2 dp)

**7. Calculate **

**a. the concentration of 32g of KOH in 500cm**^{3}** of water**

**b. the mass of iron (Fe) that can be extracted from 14.9g of iron(II) oxide (Fe**_{2}**O**_{3}**)**

**c. the volume of 100g of CO**_{2}** at normal atmospheric pressure and temperature**

**Solutions**

7a. Mr of KOH = 56

no. of mol of KOH = 32/56 = 0.5714mol --> 0.57mol (2 dp)

conc of KOH = no. of mol/(500/1000) = 1.14 mol/dm^{3}

7b. Mr of Fe_{2}O_{3} = 160

Mass of Fe extracted = (56 x 2)/160 x 14.9 = 10.43g

7c. Mr of CO_{2} = 44

no. of mol of CO_{2} = 100/44 = 2.272 mol --> 2.27 mol (2 dp)

vol of CO_{2} = 2.27mol x 24dm^{3} = 54.48dm^{3}

**8. A solution of 250cm**^{3}** of ethanol in water contained 22.3g of ethanol, C**_{2}**H5OH. Calculate the concentration of ethanol in g/dm**^{3}** and mol/dm**^{3}**.**

**Solutions**

8. Conc of ethanol in g/dm^{3} = 22.3 / (250/1000) = 89.2 g/dm^{3}

Mr of ethanol = 46

conc of ethanol in mol/dm^{3} = 89.2/46 = 1.94 mol/dm^{3}