The bivalent metal chloride weighing 9.5 g was treated with sodium hydroxide.

The bivalent metal chloride weighing 9.5 g was treated with sodium hydroxide. A precipitate with a mass of 5.8 g fell out. What metal chloride reacted?

Me Cl2 + 2 * Na OH = 2 * Na Cl + Me (OH) 2
X + 71 grams X + 34 grams
9.5 * (X + 34) = 5.8 * (X + 71)
9.5 * x + 9.5 * 34 = 5.8 * x + 5.8 * 71
9.5 * X + 323 = 5.8 * X + 411.8
We transfer the known ones to one side, and the unknown ones to the other side, when transferring values, the signs change to the opposite sign, that is, we get
9, 5 * X – 5.8 * X = 411, 8 – 323
3.7 * x = 411, 8 – 323
3.7 * X = 88.8
X = 88, 8/3, 7
x = 88.8 * 10 / (3.7 * 10)
x = 888/37
x = 24
X – Mg
Answer: Mg