Question

IA group elements tend to be reactive due to their electronic configuration ($n{s}^1$) in their valence shell. Depending on their size, they form different type of basic oxide.

57.5 kg of $Na$ is heated in air at $180{}^{o}C$. The formed pale yellow salt is allowed to stoichiometrically react with water. One of the product formed - $X$ gives no result with either red nor blue litmus paper.
Assuming 65% efficiency of the overall process, weight of $X$ is ___.

• A. 22.6 kg
• B. 24.6 kg
• C. 26.6 kg
• D. 27.6 kg

Answer 1

The correct option is D 27.6 kg

Reaction involved:
$Na+O_2⟶N{a}_{2}O...\left(i\right)$

$N{a}_{2}O+O_2\stackrel{180{}^{o}C}{\to }{\stackrel{+}{Na}}_2{\stackrel{2-}{O}}_2...\left(ii\right)$
$N{a}_2{O}_2+H_{2}O⟶\underset{\left(X\right)}{H_2{O}_2}+NaOH...\left(iii\right)$
Hydrogen peroxide does not react with litmus paper rather it bleaches it.
Moles of $Na$ taken $=\frac{57.5}{23}=2.5$

$\therefore \text{meq of}Na=\text{meq of}{H}_2{O}_2$ (n-factor of Na is 1)
Eq. (ii) is redox reaction and eq. (iii) is non-redox. From (ii) and (iii), n-factor of $H_2{O}_2$ comes out to be 2.
So equivalent of $H_2{O}_2=\frac{34}{2}=17$
Weight obtained at 65% of $H_2{O}_2$ $=2.5meq\times \frac{65}{100}\times 17g/eq=27.625kg$