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How many moles of lithium hydroxide would be required to produce 35.0 g of Li₂CO₃ in the following chemical reaction? 2 LiOH(s) + CO₂(g) → Li₂CO₃(s) + H₂O(l)

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keziakh004

Answer:

0.947 mol LiOH

Explanation:

2 LiOH  +  CO₂  ⇒  Li₂CO₃  +  H₂O

Convert grams of Li₂CO₃ to moles.  The molar mass is 73.891 g/mol.

(35.0 g Li₂CO₃)/(73.891 g/mol Li₂CO₃) = 0.4737 mol Li₂CO₃

Use the mole ratio between LiOH and Li₂CO₃ to covert moles of Li₂CO₃ to moles of LiOH.

0.4737 mol Li₂CO₃ × (2 mol LiOH/1 mol Li₂CO₃) = 0.9474 mol LiOH

Round for proper significant figures.

0.9474 ≈ 0.947 mol LiOH

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