
Calculate the amount of lime obtained by heating $400\,kg$ of limestone.
A. $56\,kg$
B. $112\,kg$
C. $224\,kg$
D. $280\,kg$
Answer
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Hint:As we all know that when limestone commonly known as calcium carbonate is heated in a limited amount of air or even in absence of air, it results in the formation of calcium oxide also called quicklime and carbon dioxide.
Complete answer:
As we all know that when limestone commonly known as calcium carbonate $(CaC{O_3})$ is heated in limited amount of air or even in absence of air, it decomposes into calcium oxide $(CaO)$ also called quicklime and carbon dioxide $C{O_2}$. Calcium oxide is yellow when it is hot but becomes white when it cools down. We can write the chemical equation for this decomposition as:
$CaC{O_3}(s) \rightleftharpoons CaO(s) + C{O_2}(s)$
We know the molecular mass of limestone is $100\,gmo{l^{ - 1}}$, molecular mass of quick lime is $56\,gmo{l^{ - 1}}$and molecular mass of carbon dioxide is $44\,gmo{l^{ - 1}}$.
From the above equation and stoichiometric calculations which are based on the stoichiometric coefficient of reactants as well as products, we can say that 1 mole of limestone produces with 1 mole of quick lime and 1 mole of carbon dioxide or in other words we can say that $100\,g$ or $100\,kg$ of limestone reacts with $56g$ or $56\,kg$ of quick lime and $44\,g$ or $44\,kg$ of carbon dioxide respectively.
$CaCO_3 \rightleftharpoons CaO+CO_2$
$1\, mole\;\;\;\;\;\;1\, mole\;\;\;\;1\, mole$
$100\,g\;\;\;\;\;\;\;\;\;56\,g\;\;\;\;\;\;\;\;44\,g$
Now, we can also say that $100\,kg$ of limestone reacts with $56\,kg$ of quick lime and $44\,kg$ of carbon dioxide, so $400\,kg$ of limestone will produces:
$ = \dfrac{{56 \times 400}}{{100}}$
$ = 224\,kg$ of quicklime.
And if we asked to calculate the amount of carbon dioxide, so $400\,kg$ of limestone will produce carbon dioxide as:
$
= \dfrac{{44 \times 400}}{{100}} \\
= 176\,kg \,of \,CO_2
$
Therefore, our correct answer is C. $224\,kg$
Note:
Using stoichiometric calculation we can identify the limiting reagents also. Limiting reagent are those chemicals which are present in lesser quantity and will be consumed first in the process of conversion of reactants into products. These chemicals can decide the amount of product formed as well as amount of reactant consumed.
Complete answer:
As we all know that when limestone commonly known as calcium carbonate $(CaC{O_3})$ is heated in limited amount of air or even in absence of air, it decomposes into calcium oxide $(CaO)$ also called quicklime and carbon dioxide $C{O_2}$. Calcium oxide is yellow when it is hot but becomes white when it cools down. We can write the chemical equation for this decomposition as:
$CaC{O_3}(s) \rightleftharpoons CaO(s) + C{O_2}(s)$
We know the molecular mass of limestone is $100\,gmo{l^{ - 1}}$, molecular mass of quick lime is $56\,gmo{l^{ - 1}}$and molecular mass of carbon dioxide is $44\,gmo{l^{ - 1}}$.
From the above equation and stoichiometric calculations which are based on the stoichiometric coefficient of reactants as well as products, we can say that 1 mole of limestone produces with 1 mole of quick lime and 1 mole of carbon dioxide or in other words we can say that $100\,g$ or $100\,kg$ of limestone reacts with $56g$ or $56\,kg$ of quick lime and $44\,g$ or $44\,kg$ of carbon dioxide respectively.
$CaCO_3 \rightleftharpoons CaO+CO_2$
$1\, mole\;\;\;\;\;\;1\, mole\;\;\;\;1\, mole$
$100\,g\;\;\;\;\;\;\;\;\;56\,g\;\;\;\;\;\;\;\;44\,g$
Now, we can also say that $100\,kg$ of limestone reacts with $56\,kg$ of quick lime and $44\,kg$ of carbon dioxide, so $400\,kg$ of limestone will produces:
$ = \dfrac{{56 \times 400}}{{100}}$
$ = 224\,kg$ of quicklime.
And if we asked to calculate the amount of carbon dioxide, so $400\,kg$ of limestone will produce carbon dioxide as:
$
= \dfrac{{44 \times 400}}{{100}} \\
= 176\,kg \,of \,CO_2
$
Therefore, our correct answer is C. $224\,kg$
Note:
Using stoichiometric calculation we can identify the limiting reagents also. Limiting reagent are those chemicals which are present in lesser quantity and will be consumed first in the process of conversion of reactants into products. These chemicals can decide the amount of product formed as well as amount of reactant consumed.
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