Drained 200 g of 20% sodium hydroxide solution and 200 g of 10% hydrochloric acid solution. Determine the mass of the salt obtained.

Given:
m (NaOH) = 200g
w (NaOH) = 20%
m (HCl) = 200g
w (HCl) = 10%
m (NaCl) -?
Decision:
NaOH + HCl = NaCl + H2O
m (HCl) = 200 * 0.1 = 20 g
n (HCl) = 20 / 36.5 = 0.55 mol
m (NaOH) = 200 * 0.2 = 40 g
n (NaOH) = 40/40 = 1 mol
According to the condition of the problem n (HCl): n (NaOH) = 1: 1, and according to the equation n (HCl): n (NaOH) = 0.55: 1
Because HCl is given in deficiency, the calculation is carried out using NaOH
n (NaCl) = n (NaOH) = 1 mol
m (NaCl) = n * M = 1 * 58.5 = 58.5 g
Answer: m (NaCl) = 1 * 58.5 = 58.5 g