0.100 mol of CaCO3 and 0.100 mol CaO are placed in an 10.0 L evacuated container and heated to 385 K. When equilibrium is reached the pressure of CO2 is 0.220 atm. 0.300 atm of CO2 is added, while keeping the temperature constant and the system is allowed to reach again equilibrium. What will be the final mass of CaCO3

12.531 grams will be the final mass of calcium carbonate.

Explanation:

Partial pressure of carbon dioxide gas at equilibrium =

Volume of carbon dioxide gas = V = 10.0 L

Moles of carbon dioxide gas formed = n

Temperature of the gas = T = 385 K

( ideal gas equation)

According to reaction , 1 mole of carbon dioxide gas is formed from 1 mole of calcium carbonate,0.06960 mole of carbon dioxide gas will be obtained from :

calcium carbonate

Moles of calcium carbonate at equilibrium = 0.100 mol - 0.06960 mol = 0.03040 mol

After addition of 0.300 atm of carbon dioxide gas, more amount of calcium carbonate will be be formed.

Here, at the same temperature, the equilibrium pressure of the carbon dioxide gas is 0.220 atm so, entire 0.300 atm of carbon dioxide will get convert to calcium carbonate.

So amount moles of carbon dioxide gas added = moles of calcium carbonate formed after re-establishment of an equilibrium :

Partial pressure of carbon dioxide gas added at equilibrium =

Volume of carbon dioxide gas = V = 10.0 L

Moles of carbon dioxide gas formed = n

Temperature of the gas = T = 385 K

( ideal gas equation)

Moles of calcium carbonate = 0.09491 mol

Total moles of calcium carbonate in the container :

=0.03040 mol + 0.09491 mol = 0.12531 mol

Mass of 0.12531 moles of calcium carbonate :

0.12531 mol × 100 g/mol = 12.531 g

12.531 grams will be the final mass of calcium carbonate.

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explanation: