Question

A 3.7gm sample of copper $\left( {ii} \right)$ carbonate is added to $25c{m^3}$
 of $2.0\,mol\,d{m^{ - 3}}$ hydrochloric acid. Which volume of gas is produced under room conditions?
A) $0.56\,d{m^3}$
B) $0.72\,d{m^3}$
C) $1.20\,d{m^3}$
D) $2.40\,d{m^3}$

Answer 1

Hint: The amount of product produce when two or more reactant take part in a chemical reaction is decided by the limiting reagent
A limiting reagent is that reactant which is present in the least amount among all reactants and thus it will decide what will be the end point of the reaction and also will be used to calculate the amount of product that is formed in the chemical reaction.
Formula used:
$moles = \dfrac{{given\,weight}}{{Molecular\,weight}}$
$M = \dfrac{n}{V}$
Here $M$ is the molarity of the compound
And $n$ is the number of moles present in a volume $V$

Complete step by step solution:
 In the given question copper $\left( {ii} \right)$ carbonate is added to hydrochloric acid and the reaction will be as follows:
$CuC{O_3} + 2HCl \to CuC{l_2} + C{O_2} + {H_2}O$
As the given question provides the given weight of copper $\left( {ii} \right)$ carbonate as $3.7gm$ we can calculate its moles using the formula given below
$moles = \dfrac{{given\,weight}}{{Molecular\,weight}}$
Using molecular weight of copper $\left( {ii} \right)$ carbonate as $123.5$
We get moles of $CuC{O_3}$ $ = \dfrac{{3.7}}{{123.5}}$
=$0.03$
Now to calculate the moles of hydrochloric acid using the formula of
$M = \dfrac{n}{V}$
Here $M$ is the molarity of the compound
And $n$ is the number of moles present in a volume $V$
Now as given the volume of Hydrochloric acid is $25c{m^3}$ and
The molarity of hydrochloric acid is given as $2.0\,mol\,d{m^{ - 3}}$
Hence after converting volume given in $c{m^3}$ into $d{m^3}$ (using $1c{m^3} = {10^{ - 3}}d{m^3}$) we can calculate the moles of Hydrochloric acid using the above formula as follow:
$2 = \dfrac{n}{{25 \times {{10}^{ - 3}}}}$
$n = 2 \times 25 \times {10^{ - 3}} = 0.05$
As we can see that the number of moles of hydrochloric acid comes less than the moles of
 Copper $\left( {ii} \right)$ carbonate it will act as a limiting reagent
As we can see that two moles of hydrochloric acid are forming one mole of $C{O_2}$
Hence $0.05$ moles of hydrochloric acid will form $\dfrac{{0.05}}{2}$ moles of $C{O_2}$
At room conditions one of mole has volume $22.4d{m^3}$
Then $\dfrac{{0.05}}{2}$ moles of $C{O_2}$ will have
Volume $ = \dfrac{{0.05}}{2} \times 22.4$
$ = 0.56\,d{m^3}$
Hence option ’A’ is the correct solution for the given question.

Note:
A limiting reagent gets completely consumed during a chemical reaction and will always decide the end point of the chemical reaction even though the other reactants have not been reduced by a small amount.
Also, the other reactants are known as excess reagents.