Redox Reactions Question 5
Question 5 - 29 January - Shift 2
The volume of $HCl$, containing $73 g L^{-1}$, required to completely neutralise $NaOH$ obtained by reacting $0.69 g$ of metallic sodium with water, is $mL$. (Nearest Integer)
(Given : molar Masses of $Na, Cl, O, H$ are 23, $35.5,16$ and $1 g mol^{-1}$ respectively)
Show Answer
Answer: (15)
Solution:
Formula: Mole Mole Analysis
Mole of $Na=\frac{0.69}{23}=3 \times 10^{-2}$
$Na+H_2 O \longrightarrow NaOH+\frac{1}{2} H_2$
By using POAC
Moles of $NaOH=3 \times 10^{-2}$
$NaOH$ reacts with $HCl$
No. of equivalent of $NaOH=$ No. of equivalent of
$HCl$
$3 \times 10^{-2} \times 1=\frac{73}{36.5} \times V($ in $L) \times 1$
$V=1.5 \times 10^{-2} L$
Volume of $HCl=15 ml$.
