In the reaction $ CaC{O_3} + 2HCl \to CaC{l_2} + {H_2}O + C{O_2} $
The volume of $ C{O_2} $ gas formed when $ 2.5{\text{g}} $ calcium carbonate are dissolved in excess hydrochloric acid at $ {0^{\text{o}}}{\text{C}} $ and $ 1{\text{ atm}} $ pressure is:
(A) $ {\text{1}}{\text{.12L}} $
(B) $ {\text{56}}{\text{.0L}} $
(C) $ {\text{0}}{\text{.28L}} $
(D) $ {\text{0}}{\text{.56L}} $
Answer
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Hint: To solve this question, you must recall basic stoichiometric fundamentals. If we know the amount of reactants in a given reaction, then we can determine the amount of products formed in the reaction. Stoichiometry is based upon the very basic laws of chemistry that help to understand it better, namely, the law of conservation of mass, the law of definite proportions (the law of constant composition), the law of reciprocal proportions and the law of multiple proportions .
Complete step by step solution:
In the question, the reaction given is $ CaC{O_3} + 2HCl \to CaC{l_2} + {H_2}O + C{O_2} $
In this reaction, we can observe that one mole of calcium carbonate in reaction with hydrochloric acid gives one mole of carbon dioxide.
Also, we know that one mole of any gas at Standard temperature and pressure conditions occupy $ {\text{22}}{\text{.4L}} $ volume
We are given that $ 2.5{\text{g}} $ calcium carbonate is reacted to.
The number of moles of Calcium carbonate are given as,
$ {{\text{n}}_{{\text{CaC}}{{\text{O}}_{\text{3}}}}} = \dfrac{{2.5}}{{100}} $
which will be equal to the number of moles of carbon dioxide produced.
Thus, the volume of carbon dioxide produced from the given amount of calcium carbonate is,
$ V = \dfrac{{2.5}}{{100}} \times 22.4 $
$ \therefore {\text{V}} = 0.56{\text{ Litres}} $
Thus, the correct answer is D.
Note:
In general, different chemicals combine in definite ratios in chemical reactions. Since matter can neither be created nor destroyed, nor can one element change into the other in a chemical reaction, thus the amount of each element must be the same throughout the entire reaction. For example, the number of atoms of any element in the reactants will be always equal to the number of atoms of that element in the products formed.
Complete step by step solution:
In the question, the reaction given is $ CaC{O_3} + 2HCl \to CaC{l_2} + {H_2}O + C{O_2} $
In this reaction, we can observe that one mole of calcium carbonate in reaction with hydrochloric acid gives one mole of carbon dioxide.
Also, we know that one mole of any gas at Standard temperature and pressure conditions occupy $ {\text{22}}{\text{.4L}} $ volume
We are given that $ 2.5{\text{g}} $ calcium carbonate is reacted to.
The number of moles of Calcium carbonate are given as,
$ {{\text{n}}_{{\text{CaC}}{{\text{O}}_{\text{3}}}}} = \dfrac{{2.5}}{{100}} $
which will be equal to the number of moles of carbon dioxide produced.
Thus, the volume of carbon dioxide produced from the given amount of calcium carbonate is,
$ V = \dfrac{{2.5}}{{100}} \times 22.4 $
$ \therefore {\text{V}} = 0.56{\text{ Litres}} $
Thus, the correct answer is D.
Note:
In general, different chemicals combine in definite ratios in chemical reactions. Since matter can neither be created nor destroyed, nor can one element change into the other in a chemical reaction, thus the amount of each element must be the same throughout the entire reaction. For example, the number of atoms of any element in the reactants will be always equal to the number of atoms of that element in the products formed.
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