chemistry help 21-2

Question

If 200.8 g of carbon dioxide gas is made from the complete combustion of isopropanol (C₃H₇OH), how many moles of oxygen are used up?

Answer 1

I love stoichiometry! We have our work cut out for us, though.

1. Write balanced reaction
2. Find relevant conversions (ex. molar mass, molar ratios)
3. Use dimensional analysis to get moles of oxygen used
4. Significant figures and rounding

1. Write balanced reaction
2. The problem tells us that this is a combustion reaction, which usually includes a fuel burning in gaseous diatomic oxygen to produce water and carbon dioxide. The fuel here is C₃H₇OH, which will influence how to balance the reaction.
3. Unbalanced: C₃H₇OH + O₂ → H₂O + CO₂
4. C balanced: C₃H₇OH + O₂ → H₂O + 3CO₂
5. H balanced: C₃H₇OH + O₂ → 4H₂O + 3CO₂
6. O is tricky to balance due to the lone O in isopropanol. I'd recommend doubling each coefficient: 2C₃H₇OH + ?O₂ → 8H₂O + 6CO₂
7. Now, we can balance O (and everything else): 2C₃H₇OH + 9O₂ → 8H₂O + 6CO₂
8. Find relevant conversions
9. We'll need to take CO₂ out of grams and put it into moles using molar mass from the periodic table:
10. Molar mass CO₂: 44.01 g CO₂ / 1 mol CO₂
11. We'll need to convert from moles of CO₂ to moles of O₂ using molar ratios from our balanced reaction:
12. Molar ratio of CO₂ to O₂: 6 mol CO₂ / 9 mol O₂
13. Use dimensional analysis to get moles of oxygen used
14. Overall process: 200.8 g CO₂ to ? mol of CO₂ to ? mol O₂
15. g CO₂ to moles CO₂: 200.8 g CO₂ * 1 mol CO₂ / 44.01 g CO₂ = 4.5626 mol CO₂
16. moles CO₂ to moles O₂: 4.5626 mol CO₂ * 9 mol O₂ / 6 mol CO₂ = 6.8439 mol O₂
17. Significant figures and rounding
18. We were given 200.8 g CO₂, which has four sig figs. We multiplied and divided throughout our dimensional analysis using quantities with no sig figs (or sig figs that we don't care about), so our answer should have four sig figs, too: 6.844 mol O₂