$2 Fe ^{2+}+ H _{2} O _{2} \rightarrow x A + y B$
(in basic medium)
$2 MnO _{4}^{-}+6 H ^{+}+5 H _{2} O _{2} \rightarrow x ^{\prime} C + y ^{\prime} D + z ^{\prime} E$
(in acidic medium)
The sum of the stoichiometric coefficients $x , y , x ^{\prime}, y ^{\prime}$ and $z ^{\prime}$ for products $A , B , C , D$ and $E ,$ respectively, is............
$\frac{ H _{2} O _{2}+2 e ^{-} \rightarrow 2 HO ^{\ominus}}{2 Fe ^{2+}+ H _{2} O _{2} \rightarrow 2 Fe ^{3+}+2 HO _{(q\omega)}^{\ominus}}$
$x =2 \quad y =2$
$\left[8 H ^{+}+ MnO _{4}^{-}+5 e ^{-} \rightarrow Mn ^{2+}+4 H _{2} O \right] \times 2$
$\left[ H _{2} O _{2} \rightarrow O _{2( g )}+2 H ^{+}+2 e ^{-}\right] \times 5$
$\Rightarrow 16 H ^{+}+2 MnO _{4}^{-}+5 H _{2} O _{2}$
$\rightarrow 2 Mn ^{2+}+8 H _{2} O +5 O _{2(g)}+10 H ^{+}$
$\Rightarrow 6 H ^{+}+2 MnO _{4}^{-}+5 H _{2} O _{2}$
$\rightarrow 2 Mn ^{2+}+8 H _{2} O +5 O _{2( g )}$
So $\quad x^{\prime}=2 \quad y^{\prime}=8 \quad z^{\prime}=5$
so $\quad x+y+x^{\prime}+y^{\prime}+z^{\prime}$
$\Rightarrow 2+2+2+8+5$
$\Rightarrow 19$
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(consider complete conversion)
$CH _{3} COOH + NaOH \rightarrow CH _{3} COONa + H _{2} O$ $(\Delta H =-55.3\,kJ\,mol ^{-1})$ The enthalpy of ionization of $CH _{3} COOH$ as calculated by the student is $kJ mol ^{-1}$. (nearest integer)
$\mathrm{Eu}^{3+}, \mathrm{Lu}^{3+}, \mathrm{Nd}^{3+}, \mathrm{La}^{3+}, \mathrm{Sm}^{3+}$