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), we add the atomic masses of its constituent elements:

• Sodium (Na) = 23 g/mol (for 2 Na: 2 × 23 = 46 g/mol)
• Carbon (C) = 12 g/mol
• Oxygen (O) = 16 g/mol (for 3 O: 3 × 16 = 48 g/mol)

Thus, the calculation is:

$ext{Relative Formula Mass} = 46 + 12 + 48 = 106 ext{ g/mol}$

First, we need to determine the moles of calcium carbonate (CaCO3) using its relative formula mass:

$ext{Moles of CaCO}_3 = \frac{\text{mass (g)}}{\text{relative formula mass}} = \frac{40000 ext{ g}}{100 ext{ g/mol}} = 400 ext{ moles}$

From the balanced equation of the Solvay process:

$2 ext{NaCl} + ext{CaCO}_3 → ext{Na}_2 ext{CO}_3 + ext{CaCl}_2$

One mole of CaCO3 produces one mole of Na2CO3. Thus, 400 moles of CaCO3 would produce 400 moles of Na2CO3.

Now, we calculate the mass of sodium carbonate (Na2CO3) that can be formed:

$\text{Mass of Na}_2 ext{CO}_3 = \text{moles} \times \text{relative formula mass} = 400 ext{ moles} × 106 ext{ g/mol} = 42400 ext{ g}$

Converting grams to kilograms:

$42400 ext{ g} = 42.4 ext{ kg}$

To find the percentage yield, we can use the formula:

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

Given that the actual yield is 104 g and the theoretical yield is 15 g, the calculation is:

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

1. Loss During Transfer: During the experimental procedure, some sodium carbonate may have been lost while transferring between containers or during filtration processes.

2. Incomplete Reactions: The reaction may not have gone to completion due to insufficient mixing, insufficient reactant quantity, or byproducts being formed that were not accounted for.