Question

NaIO3 + 5NaI + 2H3C6H5O7 → 3I2 + 3H2O + 2Na3C6H5O7

=? g/ml

Answer 1

This is a redox reaction

$5\>I^-\to\>5\>I+5e^-$

$I^{5+}+5e^-\to\>I$

Molar mass of citric acid $=192.124g/mol$

Molar mass of Iodine $=126.9g/mol$

If 2.5 of iodine is deposited by using 25ml of citric acid

Then concentration of citric acid is;

Moles of iodine deposited $=\frac{2.5}{126.9}=0.01970$

Moles of citric acid $=\frac{2}{3}×0.01970=0.01313$

Concentration of citric acid$=0.01313×\frac{1000}{25}×192.124$

$=100.9g/litre$

$=0.1009g/ml$