Question

1.42 g of a mixture of calcium carbonate and magnesium carbonate was heated to a constant weight.

The constant weight of the residue was found to be 0.76 g find out the % of magnesium carbonate

in the mixture.

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Answer 1

Answer : The percentage of magnesium carbonate is, 93.66%

Explanation :

The balanced chemical reaction will be:

$CaCO_3\rightarrow CaO+CO_2$

The mass of calcium carbonate, calcium oxide and carbon dioxide are, 100, 56 and 44 grams respectively.

As, 100 g of $CaCO_3$ react to give 56 g of $CaO$ residue

So, X g of $CaCO_3$ react to give $\frac{56\times X}{100}g$ of $CaO$ residue

The balanced chemical reaction will be:

$MgCO_3\rightarrow MgO+CO_2$

The mass of magnesium carbonate, magnesium oxide and carbon dioxide are, 84, 40 and 44 grams respectively.

As, 84 g of $MgCO_3$ react to give 40 g of $MgO$ residue

So, $1.42-Xg$ of $MgCO_3$ react to give $\frac{40\times (1.42-X)}{84}g$ of $MgO$ residue

Total mass of residue = Mass of MgO + Mass of CaO = 0.76

$\frac{56\times X}{100}+\frac{40\times (1.42-X)}{84}=0.76$

By solving the terms, we get the value of X.

X = 0.087 g

The mass of $CaCO_3$ = X = 0.087 g

The mass of $MgCO_3$ = 1.42 - X = 1.42 - 0.087 = 1.33 g

Now we have to calculate the percentage of magnesium carbonate.

$\%\text{ of }MgCO_3=\frac{\text{Mass of }MgCO_3}{\text{Total mass}}\times 100=\frac{1.33g}{1.42g}\times 100=93.66\%$

Therefore, the percentage of magnesium carbonate is, 93.66%