Question

How much quicklime can be obtained from $25$ grams of \[CaC{{O}_{3}}\] ?
A) \[18\text{ }g\]
B) \[56\text{ }g\]
C) \[14\text{ }g\]
D) \[28\text{ }g\]

Answer 1

Hint: As we all know when limestone commonly known as calcium carbonate is heated in a limited amount of the air which is even in absence of air it results in formation of calcium oxide also called quicklime as well as carbon dioxide.

Complete step-by-step answer:
As we know when limestone commonly known as calcium carbonate \[\left( CaC{{O}_{3}} \right)\] is heated in limited amount of the air, even in absence of the air, it decompose in to calcium oxide \[\left( CaO \right)\] also called quicklime as well as carbon dioxide \[\left( C{{O}_{2}} \right)\] Calcium oxide is yellow when it's hot but become white when it's cool down. We can write chemical equation for decomposition is given as:
\[CaC{{O}_{3}}_{\left( s \right)}\text{ }\to \text{ }Ca{{O}_{\left( s \right)}}\text{ }+\text{ }C{{O}_{2\left( s \right)}}\]

We know that one mole of calcium carbonate which on decomposition gives one mole of calcium oxide. Along with that the molecular weight of a calcium carbonate as well as calcium oxide is \[100\text{ }g/mol\] along with \[56\text{ }g/mol\] respectively. Thus, decomposition of a \[100\text{ }g\] of a calcium carbonate which will give us a \[56\text{ }g\] of the calcium oxide. Therefore, option C is the correct answer that is decomposition of the \[\text{25 }g\] of a calcium carbonate which will give us \[\text{14 }g\] of a calcium oxide or which we called as quicklime.

Hence the correct answer is option ‘C’.

Note: Note that using stoichiometric calculation we can identify the limiting reagent. Limiting reagent is chemical which is present in the lesser quantity as well as which will be consumed first in the process of the conversion of reactant into product. Thus, chemicals can decide the amount form and the amount of reactant consumed.