Answer:
1.57 mol; 3.14 mol
Explanation:
We have the amounts of two reactants, so this is a<em> limiting reactant problem</em>.
We know that we will need a balanced equation with moles of the compounds involved.
<em>Step 1</em>. <em>Gather all the information</em> in one place.
Mg(ClO₃)₂ + 2NaOH ⟶ Mg(OH)₂ + 2NaClO₃
<em>n</em>/mol: 2.72 3.14
<em>Step 2</em>. Identify the <em>limiting reactant</em>
Calculate the moles of Mg(OH)₂ we can obtain from each reactant.
<em>From Mg(ClO₃)₂</em>: The molar ratio of Mg(OH)₂: Mg(ClO₃)₂ is 1:1
.
Moles of Mg(OH)₂ = 2.72 × 1/1
Moles of Mg(OH)₂ = 2.72mol Mg(OH)₂
<em>From NaOH</em>: The molar ratio of Mg(OH)₂:NaOH is 1:2.
Moles of Mg(OH)₂ = 3.14 × 1/2
Moles of Mg(OH)₂ = 1.57 mol Mg(OH)₂
<em>NaOH is the limiting reactant</em> because it gives the smaller amount of Mg(OH)₂.
<em>Step 3</em>. Calculate the <em>moles of NaClO₃</em>.
The molar ratio of NaClO₃:NaOH is 2:2.
Moles of NaClO₃ = 3.14 × 2/2
Moles of NaClO₃ = 3.14 mol NaClO₃