Calculate the mass of sodium hydroxide required to react with 50g. propanoic acid.

Let’s execute the solution:
1. By the condition of the problem, we write down the equation:
m = 50 g. X g. -?
Н3С – СН2 – СООН + NaOH = H3C – CH2 – COONa + H2O – exchange, sodium propionate was obtained;
2. Calculations:
M (C2H5COOH) = 74 g / mol;
M (NaOH) = 39.9 g / mol.
3. Determine the amount of the original substance:
Y (C2H5COOH) = m / M = 50/74 = 0.675 mol;
Y (NaOH) = 0.675 mol since the amount of these substances according to the equation is 1 mol.
4. Find the mass of the base:
m (NaOH) = Y * M = 0.675 * 39.9 = 26.93 g.
Answer: to carry out the process, sodium hydroxide weighing 26.93 g is required.