This question is about oxygen (O₂) and sulfur dioxide (SO₂) - AQA - GCSE Chemistry Combined Science - Question 6 - 2019 - Paper 2

When some of the sulfur trioxide (SO₃) is removed from the reaction, the concentration of SO₃ decreases. According to Le Chatelier's principle, the equilibrium will shift to the right, towards the products, in order to favor the forward reaction and restore the concentration of SO₃. This shift helps to re-establish equilibrium.

To calculate the mass of calcium sulfite (CaSO₃) produced from 7.00 g of calcium oxide (CaO), first, we need to determine the number of moles of CaO:

1. Calculate the molar mass of CaO:

   • Molar mass of Ca = 40 g/mol
   • Molar mass of O = 16 g/mol
   • Molar mass of CaO = 40 + 16 = 56 g/mol

2. Calculate moles of CaO:

   $moles = \frac{mass}{molar \ mass} = \frac{7.00 \text{ g}}{56 \text{ g/mol}} = 0.125 \text{ moles}$

3. According to the balanced equation:

   $CaO + SO_2 \rightarrow CaSO_3$

   1 mole of CaO produces 1 mole of CaSO₃. Thus, 0.125 moles of CaO will produce 0.125 moles of CaSO₃.

4. Calculate the mass of CaSO₃ produced:

   • Molar mass of CaSO₃ = 120 g/mol (Ca = 40, S = 32, O = 16 × 3)

   $mass = moles \times molar \ mass = 0.125 \text{ moles} \times 120 \text{ g/mol} = 15.0 \text{ g}$

Therefore, the mass of calcium sulfite produced is 15.0 g.