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
