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. 0.230 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

10.32g

Explanation:

First thing's first, lets bring out the balanced chemical equation representing the reaction.

CaCO3(s) ⇄ CaO(s) + CO2(g)

From the question;

reaction constant K = [CO2(g)] = 0.220 atm.

To obtain number of moles of CO2, using ideal gas fomula, we have;

n = PV/RT = 0.220*10 / (0.08206*385) = 0.0696(mol)

From the stoichiometry of the equation above;

This is the amount of CaCO3 which has been converted to CaO before additional 0.230 atm CO2(g) is added.

After the additional 0.230 atm CO2(g), the equilibrium CO2 pressure is still 0.220 atm.  Therefore all this additional CO2 would completely convert to CaCO3:

n = PV/RT = 0.230*10.0/(0.08206*385) = 0.0728 (mol)

Hence the total CaCO3 after equilibrium is reestablished is:

0.100 - 0.0696 + 0.0728 (mol) = 0.1032 mol

Mass = Number of moles * Molar mass

Mass = 0.1032 * 100 = 10.32g

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

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