How many grams of magnesium acetate is formed when 12 g of magnesium reacts with acetic acid?

Let’s implement the solution:
We compose the reaction equation:
2CH3COOH + Mg = (CH3COO) 2Mg + H2 – substitution reaction, obtained magnesium acetate;
Let’s determine the molar masses of substances:
M (Mg) = 24.3 g / mol;
M (CH3COO) 2Mg = (12 * 2 + 3 + 16 * 2) 2 + 24.3 = 144.3 g / mol;
Let’s calculate the number of moles of magnesium, if the mass is known:
Y (Mg) = m / M = 12 / 24.3 = 0.49 mol;
Let’s make a proportion according to the reaction equation:
0.49 mol (Mg) – X mol (CH3COO) 2Mg;
-1 mol – 1 mol from here, X mol (CH3COO) 2Mg = 0.49 * 1/1 = 0.49 mol;
Let’s calculate the mass of magnesium by the formula:
M (Mg) = Y * M = 0.49 * 144.3 = 71.25 g.
Answer: the product of the reaction, magnesium acetate, was released with a mass of 71.25 g.