What volume of carbon dioxide gas at STP is required to produce 456.9g of lithium carbonate? PLEASE EXPLAIN HOW YOU GOT THE ANSWER

V =  138.68 L

Explanation:

Given data:

Volume of CO₂ at STP required = ?

Mass of lithium carbonate produced = 456.9 g

Solution;

Chemical equation:

2LiOH + CO₂     →     Li₂CO₃ + H₂O

Number of moles of Li₂CO₃ produced:

Number of moles = mass/molar mass

Number of moles = 456.9 g/ 73.89 g/mol

Number of moles = 6.184 mol

now we will compare the moles of Li₂CO₃ with CO₂ from balanced chemical equation to get number of moles of CO₂  reacted.

                      Li₂CO₃      :        CO₂

                           1           :          1

                         6.184     :        6.184

Thus, 6.184 moles of  CO₂ reacted.

Volume of CO₂:

PV = nRT

R = general gas constant = 0.0821 atm.L /mol.K

1 atm × V = 6.184 mol ×0.0821 atm.L /mol.K × 273.15 K

V = 138.68 atm.L / 1 atm

V =  138.68 L

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