Question

При действии избытка соляной кислоты на смесь карбоната и хлорида магния массой 20 г выделяется оксид углерода(4) объемом 4 л (89,3% от теоретического выхода). По этим данным определите массовую долю каждой соли в смеси.

Answer 1

MgCO3 +2HCl = MgCl2 + CO2 + H2O
Vтеор.(CO2) = 4л*100%/89,3% =4,48 л
n(CO2) = 4,48л/22,4л/моль = 0,2моль;
n(MgCO3) = n(CO2) = 0,2 моль; m(MgCO3) = 0,2моль*84г/моль=16,8 г;
W(MgCO3) = 16,8г*100%/20 г=84%; W(MgCl2) = 16%

Answer 2

Analyzing the Chemical Reaction

When an excess of hydrochloric acid acts on a mixture of magnesium carbonate and magnesium chloride with a mass of 20 g, carbon dioxide gas is produced with a volume of 4 L, which is 89.3% of the theoretical yield. We need to determine the mass fraction of each salt in the mixture.

Calculating the Mass Fraction

To calculate the mass fraction of each salt in the mixture, we can use the given data and the stoichiometry of the reaction. The balanced chemical equation for the reaction is:

MgCO3 + 2HCl → MgCl2 + CO2 + H2O

From the given data, we know that the volume of CO2 produced is 4 L and the mass of the mixture is 20 g. The molar mass of CO2 is 44 g/mol.

Using the ideal gas law, we can calculate the number of moles of CO2 produced: ``` n = PV/RT = (1 atm)(4 L)/(0.0821 L.atm/mol.K * 273 K) ≈ 0.163 moles ``` The molar ratio of MgCO3 to CO2 is 1:1, so the moles of MgCO3 consumed is also approximately 0.163 moles.

The molar mass of MgCO3 is 84 g/mol, and the molar mass of MgCl2 is 95 g/mol.

Using the molar masses and the moles of CO2 produced, we can calculate the mass of MgCO3 and MgCl2 in the mixture: ``` Mass of MgCO3 = moles of MgCO3 * molar mass of MgCO3 ≈ 0.163 moles * 84 g/mol ≈ 13.692 g

Mass of MgCl2 = (20 g - 13.692 g) (since the total mass of the mixture is 20 g) ≈ 6.308 g ```

Conclusion

The mass fraction of magnesium carbonate in the mixture is approximately 68.46% and the mass fraction of magnesium chloride is approximately 31.54%.

Note: The calculations are based on the given data and assumptions about the reaction conditions.