Mole calculation

[匿名]

2012-07-18 5:05 am

When 100g of pure CaCO3 (Formula mass =100.1) reacted with excess HCl, 44g of CO2 were obtained. However, in a similar experiment using 100g of impure Ca CO3, 46g of CO2 were obtained. Assuming that the impurity is a metallic carbonate, what would this impurity be?

(Formula masses: MgCO3 = 84.3, ZnCO3 = 125.4, FeCO3 = 115.8, BaCO3 =197.3)

A. MgCO3
B. ZnCO3
C. FeCO3
D. BaCO3

回答 (1)

fooks

2012-07-18 5:58 am

✔ 最佳答案

The answer is : A. MgCO3

CaCO3 and all the metallic carbonates in the four options have similar chemical formulae MCO3.

100 g pure CaCO3 + excess HCl → 44 g CO2 +other products
100 g impure CaCO3 + excess HCl → 46 g CO2 +other products

In case of equal masses of pure/impure CaCO3, the impure one produces more CO2.
Hence, the impure CaCO3 contains a greater number of moles of carbonates.

The total masses are both 100 g, and consider that number of moles = mass/molarmass.
The "average molar mass" of the impure CaCO3 is smaller than the molar mass of pure CaCO3.

The molar mass of pure CaCO3 is 100.1 g/mol.
Hence, the molar mass of the impurity metallic carbonate must be smaller than100.1 g/mol.
Among the four options, only option A fulfill this criterion.

參考: fooks

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