2.5 balanced equations and associated calculations

Question 1

What is the purpose of a ionic equation?

Answer 1

Ionic equations show the ions that are formed in solution and show which particles are reacting.

Question 2

What forms ions in solutions?

Answer 2

Acids, bases and salts.

Question 3

Using equations to work out masses.

Answer 3

Balanced equations can be used to work out theoretical masses. (See next card for the example).

1) Write out the equation and balance it.
2) Work out the Mr / Ar of species involved. Write these as mass in grams.
3) Divide the Ca side by 80 to find 1g then multiply by 34 to get 34g. Do the same for the CaO side.

Question 4

(Example) How much CaO can be made when 34g of Ca is burnt completely in oxygen?

Answer 4

2Ca _(g) + O_{2 (g)} → 2CaO _(s)

Ar of Ca = 40
40 x 2 = 80
80g / 80 = 1g
1g x 34 = 34g

Mr of CaO = 56
56 x 2 = 112
112g / 80 = 1.4g
1.4g x 34 = 47.6g of CaO (Theoretical mass)

Question 5

Using equations to work out volumes of gases. Balanced equations can be used to work out the volume of gases.

Answer 5

1) Write out the equation and balance it.
2) Work out the number of moles of potassium.
3) Use the equation to find out the molar ratio of K:H₂.
4) Use ideal gas equation to work out the volume.

See next card for example.

Question 6

What volume of H₂ is produced when 12g of potassium reacts with water at 100kPa of pressure and 298K? Gas constant is 8.31JK^-1mol^-1.

Answer 6

2K _(s) + 2H₂O _(l)→ 2KOH _(aq) + H_{2 (g)}

Moles = mass / Ar
moles = 12g / 39 = 0.31 moles of K

2 moles of K react to produce 1 mole of H₂
moles of H2 is 0.31 / 2 = 0.155 moles of H₂

pV = nRT but rearrange to get V = nRT / p
V = 0.155 x 8.31 x 298 / 100,000
V = 3.84 x 10^-3 m³

Question 7

Calculating percentage yield.

Answer 7

Percentage yield = actual yield / theoretical yield x 100

Question 8

What is the theoretical yield?

Answer 8

The theoretical yield is the amount of a product produced assuming no products are lost and all reactants react fully.

Question 9

(Example) In a reaction involving the complete combustion of Calcium 32.6g of Calcium Oxide was produced. The theoretical mass is 47.6g. Calculate the percentage yield of this reaction.

Answer 9

Percentage yield = actual / theoretical x 100

Percentage yield = 32.6 / 47.6 x 100 = 68.5%

Question 10

What is atom economy?

Answer 10

Atom economy is how efficient a reaction is.

Question 11

How to calculate % atom economy.

Answer 11

% atom economy = molecular mass of desired product / sum of molecular masses of all reactants x 100

Question 12

(Example) Iron oxide (Fe₂O₃) can be reduced using carbon dioxide to make pure iron and carbon dioxide. Calculate the atom economy in the extraction of iron.

Answer 12

Fe₂O₃ + 1.5C → 2Fe + 1.5CO₂

atomic mass of desired product = 2 x 55.8 = 111.6
sum of molecular masses of all reactants = (2 x 55.8) + (3 x 16.0) + (1.5 x 12.0) = 177.6

% atom economy = 111.6 / 177.6 x 100 = 62.8%

Question 13

Why is atom economy important?

Answer 13

• High atom economies produce less waste and will therefore benefit the environment.
• High atom economies mean that raw materials are used more efficiently. This is more sustainable.
• Higher atom economy means less by-products so less time and money will be spent separating these from the desirable product.
• Companies will try to use reactions that tend towards 100% atom economy.