Milk of Magnesia Question

Question

Milk of magnesia, which is an aqueous suspension of magnesium hydroxide, is used as an antacid in the reaction below. How many molecules of HCl would have to be present to form 60.38 g of MgCl₂? Mg(OH)₂(s) + 2 HCl(aq) → 2 H₂O(l) + MgCl₂(aq)

Answer 1

Hello, Izehi,

The balanced equation tells us we need 2 moles of HCL for every 1 mole of MgCl₂. If we know the number of moles of MgCl₂ in 60.38 g of the compound, we can multiply that number by 2 to tell us the number of moles of HCl reacted

Mg(OH)₂(s) + 2 HCl(aq) → 2 H₂O(l) + MgCl₂(aq)

So let's calculate the moles of MgCl₂ that are in 60.38 g of the compound. Determine the molar mass of MgCl₂. I get 95.21 g/mole.

To find moles, divide the mass by the molar mass:

60.38 grams/(95.21 g/mole) = 0.6342 moles of MgCl₂

That means we need 2*(0.6342) = 1.269 moles of HCl

The definition of a mole is 6.02 x 10²³ particles/mole (particles in this case are HCl atoms)

Therefore the number of HCl atoms required to form 60.38 grams of MgCl₂ is:

(1.269 moles HCl) * (6.02 x 10²³ atoms HCl/mole HCl) = 7.635 x 10²³ atoms of HCl.

You'd better start counting them out now. But the concept of the mole makes counting them unnecessary, since we can use moles instead and calculate the numbers of atoms we need in terms that are equivalent to grams, instead.

Bob