Question

How many grams of H20 are produced from the reaction of 30.0g of Mg(OH)2 reacting with

Answer 1

Answer: The mass of water required is 18.52 g

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Given mass of magnesium hydroxide = 30.0 g

Molar mass of magnesium hydroxide = 58.32 g/mol

Plugging values in equation 1:

$\text{Moles of magnesium hydroxide}=\frac{30.0g}{58.32g/mol}=0.5144 mol$

The given chemical equation follows:

$Mg(OH)_2+2HCl\rightarrow MgCl_2+2H_2O$

By the stoichiometry of the reaction:

If 1 mole of magnesium hydroxide produces 2 moles of water

So, 0.5144 moles of magnesium hydroxide will react with = $\frac{2}{1}\times 0.5144=1.0288mol$ of water

Molar mass of water = 18 g/mol

Plugging values in equation 1:

$\text{Mass of water}=(1.0288mol\times 18g/mol)=18.52g$

Hence, the mass of water required is 18.52 g