If 348 grams of Mg(OH)2 reacts with excess HCl (this is CL chlorine, not C and I), how many moles of MgCl2 (this is CL chlorine, not C and I) are produced?

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:

Hope this helps

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