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On the space shuttle, lithium hydroxide is produced from lithium oxide and water. The lithium hydroxide then reacts with the exhaled carbon dioxide to remove it from the air supply. The process occurs based on the following reactions: Li2O (s) + H2O (g) ĺ2LiOH (s) LiOH (s) + CO2 (g) ĺLiHCO3 (s) If you have 2.6 g of Li2O and 1.3 g of H2O, how many grams of CO2 can you remove from the air? Report your answer to two significant figures. Do NOT include units in your answer. __1__ g CO2

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Answer:

6,2g of COâ‚‚

Explanation:

Based on the reactions:

Li₂O(s) + H₂O(g) → 2LiOH(s)

LiOH(s) + CO₂(g) → LiHCO₃(s)

2,6g of Liâ‚‚O and 1,3g of Hâ‚‚O are:

2,6g × ( 1mol / 29,88g) = 0,087 moles

1.3g × ( 1mol / 18,01g) = 0,072 moles

That means limiting reactant is Hâ‚‚O. The moles produced of LiOH are:

0,072 moles of H₂O × ( 2mol LiOH / 1mol H₂O) = 0,14 moles of LiOH

Thus, the maximum COâ‚‚ that can react are 0,14 moles of COâ‚‚, in grams

0,14 moles CO₂ × (44,01g / 1mol) = 6,2g of CO₂

I hope it helps!

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