Question

The decomposition of the certain mass of $CaC{{O}_{3}}$ gave 11.2 $d{{m}^{3}}$ of $C{{O}_{2}}$ gas at STP. The mass of KOH required to completely neutralize the gas is:
A. 56g
B. 28g
C. 42g
D. 20g

Answer 1

Hint: STP stands for standard temperature and pressure. STP is most commonly used in performing the calculations on gases such as gas density. The standard temperature is 273 K, $0{}^\circ $Celsius or $32{}^\circ $Fahrenheit and the standard pressure is 1 atm pressure.

Complete answer:
The decomposition of the certain mass of $CaC{{O}_{3}}$ gave 11.2 $d{{m}^{3}}$ of $C{{O}_{2}}$ gas at STP which shows
1 mole of $C{{O}_{2}}$= 22.4 $d{{m}^{3}}$
22.4 $d{{m}^{3}}$= 44g of $C{{O}_{2}}$
11.2 $d{{m}^{3}}$= xg of $C{{O}_{2}}$
So the value of x = $\dfrac{11.2\times 44}{22.4}=22g$
The neutralization reaction is shown as:
\[\begin{align}
  & KOH+C{{O}_{2}}\to KHC{{O}_{3}} \\
 & 56\ \ \ \ \ \ \ \ \ \ \text{44} \\
\end{align}\]
56 g KOH required for neutralization of 44 g of $C{{O}_{2}}$
KOH required for neutralization of 22 g of $C{{O}_{2}}$ is calculated by
$\dfrac{56\times 22}{44}=28g$

Hence 28g KOH required to completely neutralize the gas i.e. option B is the correct answer.

Note:
Neutralization reaction generally defined as a chemical reaction in which an acid and base quantitatively react together to form a salt and water as products. These are generally acid-base neutralization reactions. These reactions generally depend on the strength of acids and bases.