The interaction of magnesium weighing 72 g with water formed a hydroxide. subtract its mass.

According to the condition of the problem, we compose the reaction equation:
Mg + 2H2O = Mg (OH) 2 + H2 – redox reaction, hydrogen gas is released;
Let’s calculate the molar masses of substances:
M (Mg) = 24.3 g / mol;
M Mg (NO3) 2 = 24.3 = 17 * 2 = 58.3 g / mol;
Determine the number of moles of magnesium:
Y (Mg) = m / M = 72 / 24.3 = 2.96 mol;
Let’s make the proportion:
2.96 mol (Mg) – X mol Mg (OH) 2;
-1 mol                -1 mol hence, X mol of Mg (OH) 2 = 2.96 * 1/1 = 2.96 mol;
Find the mass of magnesium hydroxide by the formula:
m Mg (OH) 2 = Y * M = 2.96 * 58.3 = 172.56 g.
Answer: during the reaction, magnesium hydroxide with a mass of 172.56 g is formed.