Question

You have 45.6 mL of 10.1 M HBr. You need to use Mg(OH)₂ to neutralize it. How many grams of Mg(OH)2 are

necessary?

2HBr + Mg(OH)2 → 2H₂O + MgBr₂

Answer 1

18.63 grams of Mg(OH)₂

Step-by-step explanation:

2HBr(aq) + Mg(OH)₂(aq) → MgBr₂(aq) + 2H₂O(aq)

Since the stoichiometric ratio of the above equation is 2 : 1 : 1 : 2, therefore number of moles of HBr = 2 × number of moles of Mg(OH)₂. We can use the number of moles to find the mass present. [n=m/M].

To calculate the number of moles of hydrobromic acid present in the solution, we can multiply the concentration by the volume.

n(HBr) = 0.0456×10.1 = 0.46056 mol

∴ n(Mg(OH)₂) = 1/2 n(HBr) = 1/2 × 0.46056 = 0.23028 mol.

Now we have the number of moles of magnesium hydroxide required to neutralise the hydrobromic acid, we can use this value to calculate the mass present. Molar mass can be found using a standard IUPAC Periodic Table.

m(HBr) = nM = (0.23028)(1.008+79.90) = 18.63

Therefore, we require 18.63 grams of Mg(OH)₂ to neutralise 45.6 mL of 10.1 mol/L HBr.