Question

A 12 grams of a mixture of sand and calcium carbonate on strong heating produced $7.6{\text{ grams}}$ of residue. How many grams of sand is present in the mixture?
Let mass of sand be $= x{\text{ grams}}$

Answer 1

To solve this we must first write the balanced chemical equation for decomposition of calcium carbonate. Calcium carbonate on heating decomposes and produces calcium oxide and carbon dioxide gas.to solve such questions,we should have knowledge of basic chemical formulas of molecules and formulas to be used.

Formulae used:
(1) ${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass of solute}}\left( {\text{g}} \right)}}{{{\text{Molecular mass}}\left( {{\text{g/mol}}} \right)}}$
(2) ${\text{Mass of solute}}\left( {\text{g}} \right) = {\text{Number of moles}}\left( {{\text{mol}}} \right) \times {\text{Molecular mass}}\left( {{\text{g/mol}}} \right)$

Complete step by step solution:
We know that calcium carbonate on heating decomposes and produces calcium oxide and carbon dioxide gas. The reaction is as follows:
${\text{CaC}}{{\text{O}}_3} \to {\text{CaO}} + {\text{C}}{{\text{O}}_2}$
The carbon dioxide gas escapes out. Thus, the residue that remains is of calcium oxide and sand.
We are given that $7.6{\text{ grams}}$ of residue is obtained.
Calculate the number of moles of calcium oxide using the equation as follows:
${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass of solute}}\left( {\text{g}} \right)}}{{{\text{Molecular mass}}\left( {{\text{g/mol}}} \right)}}$
Substitute $7.6{\text{ grams}}$ for the mass of calcium oxide, $56{\text{ g/mol}}$ for the molar mass of calcium oxide. Thus,
${\text{Number of moles}} = \dfrac{{7.6{\text{ grams}}}}{{56{\text{ g/mol}}}}$
${\text{Number of moles}} = 0.1357{\text{ mol}}$
Thus, the number of moles of calcium oxide are $0.1357{\text{ mol}}$.
From the reaction, we can see that one mole of calcium carbonate produces one mole of calcium oxide. Thus,
Number of moles of calcium carbonate are $0.1357{\text{ mol}}$.
Now, calculate the mass of calcium carbonate using the equation as follows:
${\text{Mass of solute}}\left( {\text{g}} \right) = {\text{Number of moles}}\left( {{\text{mol}}} \right) \times {\text{Molecular mass}}\left( {{\text{g/mol}}} \right)$
Substitute $0.1357{\text{ mol}}$ for the number of moles of calcium carbonate, $106{\text{ g/mol}}$ for the molecular mass of calcium carbonate.
Thus,
${\text{Mass of calcium carbonate}} = 0.1357{\text{ mol}} \times 106{\text{ g/mol}}$
${\text{Mass of calcium carbonate}} = 14.3842{\text{ g}}$
Thus, the mass of calcium carbonate is $14.3842{\text{ g}}$.
Now, we are given that $12{\text{ grams}}$ of mixture of calcium carbonate and sand is heated.
Thus, mass of sand is,
${\text{Mass of sand}} = \left( {14.3842 - 12} \right){\text{ g}}$
${\text{Mass of sand}} = 2.3842{\text{ g}}$

Thus, the mass of sand present in the mixture is $2.3842{\text{ g}}$.

Note: To solve this we must first correctly write the balanced chemical equation for the decomposition of calcium carbonate. This gives the correct mole ratio. The wrong or unbalanced chemical equation gives incorrect mole ratio which leads to incorrect answers.