A piece of sodium taken in excess was dipped into a solution of ethanol in benzene weighing 11.5

A piece of sodium taken in excess was dipped into a solution of ethanol in benzene weighing 11.5 grams with a mass fraction of alcohol of 0.6. Determine the mass fractions of substances in the solution after the reaction

1. Let’s write down the equation of the proceeding reaction:

2C2H5OH + 2Na → 2C2H5ONa + H2;

2.Calculate the masses of ethanol and benzene in the initial solution:

m (C2H5OH) = w (C2H5OH) * m1 (solution) = 0.6 * 11.5 = 6.9 g;

m (C6H6) = m (solution) – m (C2H5OH) = 11.5 – 6.9 = 4.6 g;

3. find the chemical amount of alcohol and sodium ethanolate:

n (C2H5OH) = m (C2H5OH): M (C2H5OH);

M (C2H5OH) = 2 * 12 + 5 + 17 = 46 g / mol;

n (C2H5OH) = 6.9: 46 = 0.15 mol;

n (C2H5ONa) = n (C2H5OH) = 0.15 mol;

4.calculate the mass of ethanolate:

m (C2H5ONa) = n (C2H5ONa) * M (C2H5ONa);

M (C2H5ONa) = 2 * 12 + 5 + 16 + 23 = 68 g / mol;

m (C2H5ONa) = 0.15 * 68 = 10.2 g;

5.sodium ethanolate and benzene will remain in the solution, find the mass of the solution:

m2 (solution) = m (C2H5ONa) + m (C6H6) = 10.2 + 4.6 = 14.8 g;

6.determine the mass fractions of substances in the resulting solution:

w (C2H5ONa) = m (C2H5ONa): m2 (solution) = 10.2: 14.8 = 0.6892 or 68.92%;

w (C6H6) = m (C6H6): m2 (solution) = 4.6: 14.8 = 0.3108 or 31.08%.

Answer: 68.92% C2H5ONa, 31.08% C6H6.