The question is missing the number of moles of carbon dioxide that will react with water.
I will work the problem with an arbitrary amount of carbon dioxide to show how to solve it.
For this, I will take 2.40 moles of carbon dioxide.
Answer:
- <u>The number of moles of oxygen atoms produced is equal to the number of moles of carbon dioixide that react: 2.40 moles of oxygen.</u>
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- 2.40 has 3 significant figures.
Explanation:
<u>1) Word equation:</u>
- carbon dioxide gas + liquid water (in the presence of light) → aqueous glucose + oxygen gas
<u>2) Balanced chemical equation</u>
- 6CO₂ (g) + 6H₂O (l) → C₆H₁₂O₆ (aq) + 6O₂(g)
<u>3) Mole ratios</u>
- 6 mol CO₂ (g) : 6 mol H₂O (l) : 1 mol C₆H₁₂O₆ (aq) : 6 mol O₂(g)
<u>4) Set a proportion:</u>
It is assumed that there is plenty liquid water (excess reactant), so you can set a proportion with the number of moles of carbon dioxide:
- 6 mol CO₂ / 6 mol O₂ = 2.40 mol CO₂ / x
From which, x = 2.40 mol O₂
So, the number of moles of oxygen produced is equal to the number of moles of carbon dioxide that react.
Since the number of moles of reactant has 3 significant figures, and the stoichiometric coefficients are considered exact, the answer also has 3 significant figures.