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in Chemistry by kratos

(I) Balance the following equations by ion electron method?

KMnO4 + SnCl2 + HCl → MnCl2 + SnCl4 + H2O + KCl

(II) Boric acid, H3 BO3 is a mild antiseptic and is often used as an eye wash. A sample contains 0.543 mol H3 BO3. What is the mass of boric acid in the sample?

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by kratos
 
Best answer

(I) KMnO4 + SnCl2 + HCl → MnCl2 + SnCl4 + H2O + KCl

Oxidation half reaction: (loss of electron)

Reduction of halfa reaction: (gain of electron)

Add H2O to balance oxygen atoms

Add Hcl to balance hydrogen atoms

KMnO4 + 5e- + 8 HCl → MnCl2 + 4H2O .......(4)

To equalize the number of electrons equation (1) × 5 and equation (2) × 2

(II) Molecular mass of H3 BO3 = (1 × 3) + (11 × 1) + (16 × 3) = 62

Boric acid sample contains 0.543 mole.

Mass of 0.543 mole of Boric acid = Molecular mass x mole

= 62 × 0.543

= 33.66 g

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