13. A mixture of MgCO3 and MgCO3.3H2O has a mass of 3.883 g. After heating to drive off all the water the mass is 2.927 g. What is the mass percent of

63.05% of MgCO3.3H2O by mass

Explanation:

of MgCO3.3H2O in the mixture?

The difference in masses after heating the mixture = Mass of water. With the mass of water we can find its moles and the moles and mass of MgCO3.3H2O to find the mass percent as follows:

Mass water:

3.883g - 2.927g = 0.956g water

Moles water -18.01g/mol-

0.956g water * (1mol/18.01g) = 0.05308 moles H2O.

Moles MgCO3.3H2O:

0.05308 moles H2O * (1mol MgCO3.3H2O / 3mol H2O) =

0.01769 moles MgCO3.3H2O

Mass MgCO3.3H2O -Molar mass: 138.3597g/mol-

0.01769 moles MgCO3.3H2O * (138.3597g/mol) = 2.448g MgCO3.3H2O

Mass percent:

2.448g MgCO3.3H2O / 3.883g Mixture * 100 =

dynamic equilibrium is the answer.