Reacting Quantities (OCR A-Level Chemistry A): Revision Notes

Example: Combustion of Hydrogen

In the combustion of hydrogen:

$2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \longrightarrow 2\text{H}_2\text{O(l)}$

The stoichiometry shows us that:

• 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water
• The mole ratio is 2:1:2

Worked Example: Calculating Mass of Aluminium Oxide

Let's calculate the mass of aluminium oxide formed when 8.10 g of aluminium completely reacts with oxygen.

Step 1: Calculate the amount of aluminium that reacts

$n(\text{Al}) = \frac{m}{M} = \frac{8.10}{27.0} = 0.300 \text{ mol}$

Step 2: Use the balanced equation to find the amount of Al₂O₃ produced

From the equation, 4 moles of Al produce 2 moles of Al₂O₃

Therefore: 0.300 mol of Al produces 0.150 mol of Al₂O₃

Step 3: Calculate the mass of Al₂O₃ formed

$n(\text{Al}_2\text{O}_3) = \frac{m}{M}$

Rearranging: $m = n \times M$

$M_r(\text{Al}_2\text{O}_3) = (27.0 \times 2) + (16.0 \times 3) = 102.0$

$m = 0.150 \times 102.0 = 15.3 \text{ g}$

Worked Example: Mass, Gas Volumes, and Concentration

This example shows how to calculate both concentration and gas volume from a reaction. Consider the reaction: 0.552 g of lithium reacts with water to form 125 cm³ of lithium hydroxide solution and hydrogen gas.

Step 1: Calculate the amount of lithium that reacts

$n(\text{Li}) = \frac{m}{M} = \frac{0.552}{6.9} = 0.0800 \text{ mol}$

Step 2: Use the equation to find the amounts of LiOH and H₂ formed

From the equation:

• 0.0800 mol of Li produces 0.0800 mol of LiOH
• 0.0800 mol of Li produces 0.0400 mol of H₂

Step 3: Calculate the concentration of LiOH(aq) and volume of H₂(g)

For concentration of LiOH:

$n = c \times \frac{V}{1000}$

Rearranging: $c = \frac{1000 \times n}{V} = \frac{1000 \times 0.0800}{125} = 0.640 \text{ mol dm}^{-3}$

For volume of H₂ at RTP (room temperature and pressure):

$n = \frac{V}{24000}$

Rearranging: $V = n \times 24000 = 0.0400 \times 24000 = 960 \text{ cm}^3$

Worked Example: Identifying an Unknown Group 2 Metal

If 0.14 g of an unknown Group 2 metal X produces 84 cm³ of hydrogen:

Step 1: Calculate moles of H₂ produced

$n(\text{H}_2) = \frac{V}{24000} = \frac{84}{24000} = 0.00350 \text{ mol}$

Step 2: Use the equation to find moles of metal X

$\text{X(s)} + 2\text{HCl(aq)} \longrightarrow \text{XCl}_2\text{(aq)} + \text{H}_2\text{(g)}$

From the 1:1 mole ratio: 0.00350 mol of H₂ means 0.00350 mol of X reacted

Step 3: Calculate the molar mass

$M(\text{X}) = \frac{m}{n} = \frac{0.14}{0.00350} = 40 \text{ g mol}^{-1}$

From the periodic table, calcium has a relative atomic mass of 40.1, so X is calcium.

Worked Example: Percentage Yield Calculation

Calculate the percentage yield when 1.15 g of sodium reacts with excess chlorine to form 1.872 g of sodium chloride.

Step 1: Calculate moles of sodium

$n(\text{Na}) = \frac{m}{M} = \frac{1.15}{23.0} = 0.0500 \text{ mol}$

Step 2: Calculate theoretical yield of NaCl in moles

From the 2:2 mole ratio, 0.0500 mol of Na should produce 0.0500 mol of NaCl (theoretical yield)

Step 3: Calculate actual yield of NaCl in moles

$n(\text{NaCl}) = \frac{m}{M} = \frac{1.872}{58.5} = 0.0320 \text{ mol}$

Step 4: Calculate percentage yield

$\text{percentage yield} = \frac{0.0320}{0.0500} \times 100 = 64.0\%$

Example: Hydrogen and Oxygen Reaction

When hydrogen and oxygen react to form water:

$2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \longrightarrow 2\text{H}_2\text{O(l)}$

The equation shows that 2 moles of H₂ are required for every 1 mole of O₂.

If equal molar amounts of hydrogen and oxygen are allowed to react:

• Hydrogen will be used up completely first
• Half the oxygen will remain unreacted
• Hydrogen is the limiting reagent

Worked Example: Atom Economy for Hydrogen Production

Hydrogen is produced industrially from the reaction of carbon with steam:

$\text{C(s)} + 2\text{H}_2\text{O(g)} \longrightarrow 2\text{H}_2\text{(g)} + \text{CO}_2\text{(g)}$

Step 1: Write the equation and calculate molar masses

Products:

• Desired product: 2H₂ with total molar mass = 2 × 2.0 = 4.0
• All products: 2H₂ + CO₂ with total molar mass = 4.0 + 44.0 = 48.0

Step 2: Calculate atom economy

$\text{atom economy} = \frac{2 \times 2.0}{2 \times 2.0 + 44.0} \times 100 = \frac{4.0}{48.0} \times 100 = 8.3\%$