Chemistry 0620 · IGCSE · Reversible reactions and equilibrium

Reversible reactions and equilibrium — practice question

This question concerns sulfur and sulfur compounds. Sulfur is changed into sulfuric acid, H$_2$SO$_4$, by the Contact process. The process has four stages: stage 1 Molten sulfur is changed into sulfur dioxide. stage 2 Sulfur dioxide reacts with oxygen to make sulfur trioxide. stage 3 Sulfur trioxide joins with concentrated sulfuric acid to produce oleum, H$_2$S$_2$O$_7$. stage 4 Oleum reacts to produce concentrated sulfuric acid.
(a(i))[1]

Balance the equation for the reaction when iron pyrites reacts with oxygen in air. [BLANK]FeS$_2$ + [BLANK]O$_2$ $\rightarrow$ [BLANK]Fe$_2$O$_3$ + [BLANK]SO$_2$

(a(ii))[1]

State the name of Fe$_2$O$_3$. Give the oxidation number of iron.

(b(i))[1]

Ignoring cost, explain why a temperature above $450^\circ\text{C}$ is not chosen.

(b(ii))[1]

Explain why a pressure below $2\,$atm is not chosen.

(c)[2]

Write the symbol equation for the reaction.

(d(i))[1]

Name a solution that may be mixed with aqueous ammonium sulfate to make a precipitate of lead(II) sulfate.

(d(ii))[3]

Write an ionic equation for this precipitation reaction. Add state symbols.

(d(iii))[3]

Describe how pure lead(II) sulfate may be obtained from the mixture.

Worked solution & mark scheme

This 13-mark question has a full step-by-step worked solution and mark scheme. One marking point: Balanced equation: 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂

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