(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?

(i) Balnce the following equations by ion electron method. KMnO_(4) +

1.	(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?
Answer» Solution :(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