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 sodium carbonate (Na₂CO₃), sum the relative atomic masses of its constituent elements:

• Sodium (Na): 23 (2 atoms) → 2 × 23 = 46
• Carbon (C): 12 (1 atom) → 12
• Oxygen (O): 16 (3 atoms) → 3 × 16 = 48

Thus, the relative formula mass of Na₂CO₃ is:

$46 + 12 + 48 = 106$

From the equation of the reaction:

$2NaCl + CaCO₃ → Na₂CO₃ + CaCl₂$

1 mole of CaCO₃ produces 1 mole of Na₂CO₃. The molar mass of CaCO₃ is 100 g. Therefore, the number of moles in 40 kg (40000 g) of CaCO₃ is given by:

$ext{moles of CaCO₃} = \frac{40000 ext{ g}}{100 ext{ g/mol}} = 400 ext{ moles}$

Since the reaction produces an equal number of moles of Na₂CO₃, we also obtain 400 moles of Na₂CO₃. To find the mass of sodium carbonate produced:

$ext{mass} = ext{moles} × ext{molar mass} = 400 ext{ moles} × 106 ext{ g/mol} = 42400 ext{ g} = 42.4 ext{ kg}$

To calculate the percentage yield of sodium carbonate, use the formula:

$ext{Percentage Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100$

Here, the actual yield is 104 g and the theoretical yield is 15.0 g:

$\text{Percentage Yield} = \left( \frac{104 ext{ g}}{15.0 ext{ g}} \right) \times 100 \approx 693.33\%$

1. Incomplete Reactions: The reaction may not have gone to completion, leaving unreacted reactants and leading to a lower yield.

2. Loss during Transfer: Some sodium carbonate may have been lost during the transfer between containers or during the filtration process, affecting the final actual yield.