Carbon dioxide reacts with calcium hydroxide according to the following reaction
ID: 919849 • Letter: C
Question
Carbon dioxide reacts with calcium hydroxide according to the following reaction: CO2(g) +Ca(OH)2 (aq) = CaCo(s) + H2O(l) a 5.0 L container contains 1.0 L Ca(OH)2 solution. CO2 (g) is introduced to the container. At the beginning the CO2 pressure is 4.0 atm. When the reaction is complete, the pressure drops to zero. The temperature is 298K. Assume CO2 behaves as an ideal gas. From this information calculate 1) the extent of the reaction, and 2) the number of grams CaCO3 (s) formed Carbon dioxide reacts with calcium hydroxide according to the following reaction: CO2(g) +Ca(OH)2 (aq) = CaCo(s) + H2O(l) a 5.0 L container contains 1.0 L Ca(OH)2 solution. CO2 (g) is introduced to the container. At the beginning the CO2 pressure is 4.0 atm. When the reaction is complete, the pressure drops to zero. The temperature is 298K. Assume CO2 behaves as an ideal gas. From this information calculate 1) the extent of the reaction, and 2) the number of grams CaCO3 (s) formed Carbon dioxide reacts with calcium hydroxide according to the following reaction: CO2(g) +Ca(OH)2 (aq) = CaCo(s) + H2O(l) a 5.0 L container contains 1.0 L Ca(OH)2 solution. CO2 (g) is introduced to the container. At the beginning the CO2 pressure is 4.0 atm. When the reaction is complete, the pressure drops to zero. The temperature is 298K. Assume CO2 behaves as an ideal gas. From this information calculate 1) the extent of the reaction, and 2) the number of grams CaCO3 (s) formedExplanation / Answer
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