In: Chemistry
1)
Since all of CO2 has reacted, the reaction is fully complete
Hence the extent of the reaction is 100 % complete.
2)
Lets calculate the number of moles of CO2 present initially.
Volume occupied by CO2 = 4 L
usE:
P*V = n*R*T
4 atm * 4 L = n * 0.0821 atm L/mol-K * 298
n = 0.654 mol
So, 0.654 mol of CO2 has reacted.
From the reaction:
CO2(g) +Ca(OH)2 (aq) = CaCo3(s) + H2O(l)
1 mol of CO2 gives 1 mol of CaCO3.
So, number of moles of CaCo3 formed = 0.654 mol
Molar mass of CaCo3= 100 g/mol
So,
mass of CaCo3 formed = number of moles * molar mass
= 0.654 * 100
= 65.4 g
Answer: 65.4 g