Sulfuric acid and an excess of sodium iodide solution were added to a solution of sodium dichromate weighing 250 g.

Sulfuric acid and an excess of sodium iodide solution were added to a solution of sodium dichromate weighing 250 g. In this case, iodine with a mass of 15.24 g was formed. Calculate the mass fraction of sodium dichromate in the original solution.

1. Let’s write down the reaction equation:

Na2Cr2O7 + 6NaI + 7H2SO4 = 3I2 + 4Na2SO4 + Cr2 (SO4) 3 + 7H2O;

2.Calculate the chemical amount of iodine:

n (I2) = m (I2): M (I2) = 15.24: 254 = 0.06 mol;

3. find the amount of sodium dichromate:

n (Na2Cr2O7) = n (I2): 3 = 0.06: 3 = 0.02 mol;

4.calculate the mass of the dichromate:

m (Na2Cr2O7) = n (Na2Cr2O7) * M (Na2Cr2O7);

M (Na2Cr2O7) = 2 * 23 + 2 * 52 + 7 * 16 = 262 g / mol;

m (Na2Cr2O7) = 0.02 * 262 = 5.24 g;

5.determine the mass fraction of Na2Cr2O7 in the initial solution:

w (Na2Cr2O7) = m (Na2Cr2O7): m (solution);

w (Na2Cr2O7) = 5.24: 250 = 0.02096 or 2.096%.

Answer: 2.096%.