Sulfur dioxide is the desired product in two possible reaction pathways.
Route 1:
S+O2→SO2\text{S} + \text{O}_2 \rightarrow \text{SO}_2S+O2→SO2
Route 2:
4FeS2+11O2→2Fe2O3+8SO24\text{FeS}_2 + 11\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 + 8\text{SO}_24FeS2+11O2→2Fe2O3+8SO2
Relative atomic masses: Fe=56\text{Fe} = 56Fe=56, S=32\text{S} = 32S=32, O=16\text{O} = 16O=16.
Trials of the two routes gave these results.
| Route | Percentage yield in % | Time to produce 1.0 tonne of sulfur dioxide in hours |
|---|---|---|
| 1 | 72 | 5.0 |
| 2 | 94 | 3.0 |
Calculate the percentage atom economy for the production of sulfur dioxide by each route.
Evaluate the two routes and suggest which route should be used.
104 exam-style questions on AQA GCSE Chemistry 3.3 Yield and atom economy of chemical reactions, covering 3.3.1a Percentage yield, 3.3.1b Theoretical mass of a product, 3.3.2a Atom economy, and 3.3.2b Choosing a reaction pathway. Each one has a worked solution and a mark scheme showing where the marks go.