An excess of sodium hydroxide was added to a solution of iron (II) chloride with a mass fraction of 5%

An excess of sodium hydroxide was added to a solution of iron (II) chloride with a mass fraction of 5%; as a result of the reaction, a precipitate with a mass of 4.5 g was formed. Determine the mass of the original salt solution.

Given:
ω (FeCl2) = 5%;
m (Fe (OH) 2) = 4.5 g;
m solution (FeCl2) -?
Decision:
We will use the following formulas:
n = m / M;
ω = m in-va / m solution.
Reaction equation:
FeCl2 + 2NaOH = Fe (OH) 2 ↓ + 2NaCl.
M (Fe (OH) 2) = 56 + 17 * 3 = 90 (g / mol).
n (Fe (OH) 2) = 4.5 / 90 = 0.05 (mol).
Taking into account the ratio of the amount of Fe (OH) 2 and FeCl2 in the equation, we get that:
n (FeCl2) = 0.05 (mol).
m (FeCl2) = n (FeCl2) * M (FeCl2) = 0.05 * 127 = 6.35 (g).
m solution (FeCl2) = 6.35 / 0.05 = 127 (g).
Answer: the mass of the initial solution is 127 g.