Question

The decomposition of a certain mass of $\mathrm{CaCO}_{3}$ gave 11.2 $ dm^3 $of $\mathrm{CO}_{2}$
 gas at STP. The mass of KOH required to completely neutralise the gas is:
A. 56g
B. 28g
C. 42g
D. 20g

Answer 1

Hint: We will find the stoichiometric coefficient of the compounds and solve the numerical using the cross multiplication formulae. The numerical is based on the concepts of the chapter of class 12, that is, "some basic concepts of chemistry."

Complete step by step answer:
The stoichiometric coefficient of $\mathrm{CaCO}_{3}$ is $\dfrac{1}{2}$ and of KOH is 1. That is, half a mole of $\mathrm{CaCO}_{3}$ is equal to one mole of KOH. We need to find the neutralizing amount of $\mathrm{CO}_{2}$ required for the complete neutralization of the reaction.
Volume of $\mathrm{CO}_{2}$= 11.2 $ dm^3 $
For finding the number of moles, we take the given number of moles and divide it by the total number of moles.
We have,
1 mole of $\mathrm{CO}_{2}$ = 22.4 $ dm^3 $
$\Rightarrow $44 gm of of $\mathrm{CO}_{2}$ = 22.4 $ dm^3 $
$\Rightarrow $x gm of $\mathrm{CO}_{2}$ = 11.2 $ dm^3 $
$\Rightarrow $x =$\dfrac{11.2 \times 44}{22.4}$
$\Rightarrow x\, =22 \mathrm{gm}$
The neutralisation reaction is as,
$ KOH \, + \,CO_{2} \rightarrow KHCO_{3} $
56 gm KOH required for neutralisation of 44 gm of $\mathrm{CO}_{2}$
KOH required for neutralisation of 22 gm of $\mathrm{CO}_{2}$
$= {\dfrac{56 \times 22}{44}}$
$=28 \mathrm{gm}$
28 gm of KOH is required for complete neutralisation of 22 gm of $\mathrm{CO}_{2}$

Therefore the correct answer is option B. 28g

Note:
Focus on the units in which the numbers are given as the units' conversion is quite necessary. If you miss the conversion, your answer can go wrong, so we need to focus on the units in the questions.