In industry sodium carbonate is made from sodium chloride solution and calcium carbonate in the Solvay Process - Edexcel - GCSE Chemistry - Question 3 - 2013 - Paper 1

To calculate the relative formula mass of calcium chloride (CaCl₂), use the relative atomic masses:

• Calcium (Ca): 40
• Chlorine (Cl): 35.5

The formula mass can be calculated as:

$\text{Relative Formula Mass of CaCl}_2 = \text{mass of Ca} + 2 \times \text{mass of Cl} = 40 + 2 \times 35.5 = 40 + 71 = 111$

Using the balanced equation:

$2NaCl + CaCO_3 → Na_2CO_3 + CaCl_2$

1. Calculate Moles of Reactants:

   • For CaCO₃:

   Given mass = 40 kg = 40000 g.

   Moles of CaCO₃ = (rac{\text{mass}}{\text{molar mass}} = \frac{40000}{100} = 400 ext{ moles})

2. Determine Moles of Na₂CO₃ Produced:

   • According to the equation, 1 mole of CaCO₃ produces 1 mole of Na₂CO₃. Thus, 400 moles of CaCO₃ produce 400 moles of Na₂CO₃.

3. Calculate Mass of Sodium Carbonate:

   Molar mass of Na₂CO₃ = 106 g

   Total mass produced = moles × molar mass = 400 moles × 106 g = 42400 g = 42.4 kg.

Therefore, the maximum mass of sodium carbonate that could be formed is 42.4 kg.

To calculate the percentage yield, use the formula:

$\text{Percentage Yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100$

Given:

• Actual yield = 104 g
• Theoretical yield = 15.0 g

Substituting the values:

$\text{Percentage Yield} = \left( \frac{104}{15} \right) \times 100 \approx 693.33 \%$

1. Incomplete Reactions: Some reactants may not have fully reacted, leading to less product formation than expected.

2. Loss of Product During Transfer: Sodium carbonate may have been lost during the transfer process or while filtering, hence reducing the final yield.