(i) Balnce the following equations by ion electron method. `KMnO_(4) + SnCl_(2) + HCl rarr MnCl_(2) + SnCl_(4) + H_(2)O + KCl`
(ii) Boric acid, `H_(3)BO_(3)` is a mild antiseptic and often used as eye wash. A sample contains 0.543 mol `H_(3)BO_(3)`. What is the mass of boric acid in the sample?

Explanation

(i) `KMnO_(4) + SnCl_(2) + HCl rarr MnCl_(2) + SnCl_(4) + H_(2)O + KCl`
Oxidation half reaction: (loss of electrons)
`overset(+2)(SnCl_(2)) rarr overset(+4)(SnCl_(4)) + 2e^(-)` ....(1)
Reduction half reaction: (gain of electrons)
`overset(+7)(KMnO_(4)) + 5e^(-) rarr overset(+2)(MnCl_(2))` ....(2)
Add `H_(2)O` to balance oxygen atoms.
`KMnO_(4) + 5e^(-) rarr overset(+2)(MnCl_(2)) + 4H_(2)O` ....(3)
Add HCl to balance hydrogen atoms.
`KMnO_(4) + 5e^(-) + 8HCl rarr MnCl_(2) + 4H_(2)O` ....(4)
To equalize the number of electrons equation `(1) xx 5` and equation `(2) xx 2`
`5SnCl_(2) rarr 5SnCl_(4) + cancel(10e)`
`2KMnO_(4) + 16HCl + cancel(10e) rarr 2MnCl_(2) + 4H_(2)O + 2KCl`
`2KMnO_(4) + 5SnCl_(2) + 16HCl rarr 5SnCl_(4) + 2MnCl_(2) + 4H_(2)O + 2KCl`
(ii) Molecular mass of `H_(3)BO_(3)` = `(1 xx 3) + (11 xx 1) + (16 xx 3) = 62`
Boric acid sample contains 0.543 mole.
Mass of 0.543 mole of Boric acid = Molecular mass `xx` mole
= `62 xx 0.543` = 33.66 g