Question

How many moles of water would form the reaction of exactly 58.3 grams of magnesium hydroxide

Answer 1

Answer:

$\boxed{\text{2.00 mol}}$

Explanation:

We know we will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.

You don't tell us what the reaction is, but we can solve the problem so long as we balance the OH.

M_r:      58.32

          Mg(OH)₂ + … ⟶ … + 2HOH

m/g:       58.3

(a) Moles of Mg(OH)₂

$\text{Moles of Mg(OH) _{2} } =\text{58.3 g Mg(OH) _{2} } \times \dfrac{\text{1 mol Mg(OH) _{2} }}{\text{58.32 g Mg(OH) _{2} }}\\\\=\text{0.9997 mol Mg(OH) _{2} }$

(b) Moles of H₂O

The molar ratio is 2 mol H₂O = 1 mol Mg(OH)₂.

$\text{Moles of H _{2} O}= \text{0.9995 mol Mg(OH) _{2} } \times \dfrac{\text{2 mol {H _{2} O}}}{ \text{1 mol Mg(OH) _{2} }}\\\\= \textbf{2.00 mol H _{2} O}$

The reaction will form $\boxed{\textbf{2.00 mol}}$ of water.