Of 0.605 g of magnesium hydroxide reacts with 0.800 g of sulfuric acid, what is the mass of magnesium sulfate produced?
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Mass = 0.963 g

Explanation:

Given data:

Mass of Mg(OH)₂ = 0.605 g

Mass of H₂SO₄ = 0.800 g

Mass of MgSO₄ = ?

Solution:

Chemical equation:

Mg(OH)₂  +  H₂SO₄   →    MgSO₄ + 2H₂O

First of all we will calculate the number of moles of each reactant.

Number of moles of Mg(OH)₂:

Number of moles = mass / molar mass

Number of moles = 0.605 g / 58.32 g/mol

Number of moles = 0.01 mol

Number of moles of H₂SO₄ :

Number of moles = mass / molar mass

Number of moles = 0.800g / 98.08 g/mol

Number of moles = 0.008 mol

Now we will compare the moles of MgSO₄ with  Mg(OH)₂ and  H₂SO₄  

         Mg(OH)₂         :           MgSO₄    

               1                 :               1

              0.01            :            0.01

             H₂SO₄             :           MgSO₄    

               1                      :               1

              0.008                :            0.008

The number of moles of magnesium sulfate produced by sulfuric acid are less so it will be a limiting reactant.

Mass of magnesium sulfate:

Mass = number of moles × molar mass

Mass = 0.008 mol × 120.37 g/mol

Mass = 0.963 g

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