Percentage of tin in an alloy??

Question

Balance the following equations

KI03 + KI + H2S04 -> I2 + K2SO4 + H20

SnCl2 + I2 -> SnI4 + SnCl4

The percentage of tin in an alloy was determined by the following experimental procedure 1.5g of the alloy was dissolved in concentrated sulphuric acid, This was diluted with deionised water and concentrated hydrochloric acid was the added. The solution was boiled, cooled under a reducing atmosphere and then titrated with a standard Potassium Iodate-Potassium Iodide solution, the potassium iodide was in excess, If the concentration of the potassium iodate solution was 3.26g dm^-3 and 24.5 cm^3 of it were required in the titration calculate the percentage of tin in the alloy

Answer 1

Actually

KIO3 + 5 KI + 3 H2SO4 >> 3 I2 + 3 H2O + 3 K2SO4

2 SnCl2 + 2 I2 >> SnCl4 + SnI4

Moles KIO3 = 3.26 g /214.00 g/mol = 0.0152

M = 0.0152 / 1.00L = 0.0152

moles KIO3 = 0.0245 x 0.0152 =0.000372

moles I2 produced = 0.000372 x 3 =0.00112

moles SnCl2 =0.00112

mass Sn = 0.00112 x 118.69 = 0.133 g

% Sn = 0.133 x 100 / 1.5 =8.86

Answer 2

KIO3 + 5 KI + 3 H2SO4 >> 3 I2 + 3 H2O + 3 K2SO4

2 SnCl2 + 2 I2 >> SnCl4 + SnI4

Moles KIO3 = 3.26 g /214.00 g/mol = 0.0152

M = 0.0152 / 1.00L = 0.0152

moles KIO3 = 0.0245 x 0.0152 =0.000372

moles I2 produced = 0.000372 x 3 =0.00112

moles SnCl2 =0.00112

mass Sn = 0.00112 x 118.69 = 0.133 g

% Sn = 0.133 x 100 / 1.5 =8.86