Answer:
moles MgCl₂ = 5.97 moles
Explanation:
First, let0s write the overall reaction that is taking place here:
Mg(OH)₂ + HCl ----------> MgCl₂ + H₂O
Now, we need to balance this reaction:
Mg(OH)₂ + 2HCl ----------> MgCl₂ + 2H₂O
Now, the exercise already stated that the HCl is in excess, so we can assume the magnesium hydroxide is the limiting reactant. If this is true, then, to get the moles of MgCl₂ produced, all we need to do is calculate the moles of Mg(OH)₂ that reacted, using the following expression:
moles = mass / MM
The atomic weights for Mg, O and H are:
Mg: 24.305 g/mol; H: 1 g/mol; O = 15.999 g/mol
Let's determine the molar mass of Mg(OH)₂:
MM = 24.305 + (2 * 15.999) + (2 * 1) = 58.303 g/mol
Now, let's determine the moles:
moles = 348 g / 58.303 g/mol = 5.97 moles of Mg(OH)₂
As we can see in the overall reaction, there is a mole ration between Mg(OH)₂ and MgCl₂ of 1:1, that's why we can conclude that the moles of Mg(OH)₂ would be the same moles of MgCl₂:
<h2>
moles Mg(OH)₂ = moles MgCl₂ = 5.97 moles</h2>
Hope this helps
