Calculate the amount of sodium phenolate substance formed by the interaction of 1.4 kg of phenol with 1.4 kg of sodium hydroxide.

To solve the problem, let’s compose the reaction equation:
С6Н5ОН + NaOH = C6H5ONa + H2O – ion exchange reaction, sodium phenolate was obtained;
Let’s make calculations using the formulas:
M (C6H5OH) = 12 * 6 + 1 * 6 + 16 = 94 g / mol;
M (NaOH) = 22.9 + 1 + 16 = 39.9 g / mol;
M (C6H5ONa) = 12 * 6 + 1 * 5 + 16 + 22.9 = 43.9 g / mol;
We calculate the amount of moles of phenol, sodium hydroxide by the formula:
Y (C6H5OH) = m / M = 1400/94 = 1 4.89 mol (deficient substance);
Y (NaOH) = m / M = 1400 / 39.9 = 35.08 g;
Let’s make the proportion:
14.89 mol (C6H5OH) – X mol (C6H5OHa);
-1 mol                            -1 mol from here, X mol (C6H5OHa) = 14.89 * 1/1 = 14.89 mol;
We find the mass of sodium phenolate by the formula:
m (C6H5ONa) = Y * M = 14.89 * 43.9 = 653.67 g.
Answer: in the course of the reaction, sodium phenolate weighing 653.67 g is formed.