Answer:
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 =
