A $50{\text{ g}}$ of an impure calcium carbonate sample decomposes on heating to give carbon dioxide and $22.4{\text{ g}}$ calcium oxide. The percentage purity of calcium carbonate in the sample is:
A.$60\% $
B.$80\% $
C.$90\% $
D.$70\% $
Answer
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Hint:A decomposition reaction is the reaction which involves the breaking of a chemical compound by the action of heat, light or electrical energy to give two or more new substances. A decomposition reaction is caused when heat is provided to the compound is known as thermal decomposition reaction.
Complete step by step answer:
When heated, calcium carbonate like many other metal carbonates undergoes thermal decomposition reaction giving calcium oxide, also known as quicklime, and carbon dioxide gas. This reaction is known as calcination and is given as,
$CaC{O_3}\xrightarrow[{}]{\Delta }CaO + C{O_2}$
From the reaction we can see that 1 mole of calcium carbonate gives 1 mole of calcium oxide.
Or, $100{\text{g}}$ of calcium carbonate will give $56{\text{g}}$ of calcium oxide.
Thus, $50{\text{g}}$ of calcium carbonate must give $28{\text{g}}$ of calcium oxide but instead the calcium oxide obtained is $22.4{\text{g}}$
Hence, the amount of calcium carbonate in $50{\text{g}}$ of the impure sample is $x = \dfrac{{22.4}}{{56}}{\kern 1pt} \times 100 = 40{\text{g}}$
So the percentage purity of the given sample can be calculated as, $\% {\text{purity}} = \dfrac{{40}}{{50}} \times 100$
$\% {\text{purity}} = 80\% $
Hence, the correct answer is B.
Note:
Calcium carbonate is a chemical compound which has the chemical formula ${\text{CaC}}{{\text{O}}_{\text{3}}}$. It is found in minerals and rocks like calcite and aragonite and is commonly known as limestone. It comprises the majority of the pearls and shells of marine organisms, eggs and snails. It is used as a dietary supplement as a source of calcium as the compound is ionic and breaks into calcium ion and carbonate ion. It is a mildly basic salt and thus used in mild neutralization reactions like those required to maintain the pH of soil, or those involving its use as an antacid in humans.
Complete step by step answer:
When heated, calcium carbonate like many other metal carbonates undergoes thermal decomposition reaction giving calcium oxide, also known as quicklime, and carbon dioxide gas. This reaction is known as calcination and is given as,
$CaC{O_3}\xrightarrow[{}]{\Delta }CaO + C{O_2}$
From the reaction we can see that 1 mole of calcium carbonate gives 1 mole of calcium oxide.
Or, $100{\text{g}}$ of calcium carbonate will give $56{\text{g}}$ of calcium oxide.
Thus, $50{\text{g}}$ of calcium carbonate must give $28{\text{g}}$ of calcium oxide but instead the calcium oxide obtained is $22.4{\text{g}}$
Hence, the amount of calcium carbonate in $50{\text{g}}$ of the impure sample is $x = \dfrac{{22.4}}{{56}}{\kern 1pt} \times 100 = 40{\text{g}}$
So the percentage purity of the given sample can be calculated as, $\% {\text{purity}} = \dfrac{{40}}{{50}} \times 100$
$\% {\text{purity}} = 80\% $
Hence, the correct answer is B.
Note:
Calcium carbonate is a chemical compound which has the chemical formula ${\text{CaC}}{{\text{O}}_{\text{3}}}$. It is found in minerals and rocks like calcite and aragonite and is commonly known as limestone. It comprises the majority of the pearls and shells of marine organisms, eggs and snails. It is used as a dietary supplement as a source of calcium as the compound is ionic and breaks into calcium ion and carbonate ion. It is a mildly basic salt and thus used in mild neutralization reactions like those required to maintain the pH of soil, or those involving its use as an antacid in humans.
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