How many grams of MgCl2 will be produced if you have 15 g of Mg(OH)2 and 25 g of HCl?

Question

Can you please answer these? Thank you

Answer 1

First, write the correctly balanced equation:

Mg(OH)₂ + 2HCl ==> MgCl₂ + 2H₂O ... balanced equation

Next, since we are provided with the mass of each reactant, we must find which one is present in limited supply. An easy way to do this is to divided the moles of each reactant by the respective coefficient in the balanced equation, and whichever value is less is the limiting reactant. Thus...

For Mg(OH)₂: 15 g x 1 mol Mg(OH)₂ / 58.3 g = 0.257 mols Mg(OH)₂ ... (÷1->0.257)

For HCl: 25 g x 1 mol HCl / 36.5 g = 0.685 mols HCl ... (÷2->0.342)

Since 0.257 is less than 0.342, Mg(OH)₂ is the limiting reactant and will determine the amount of MgCl₂ that can be produced (see below).

Theoretical yield of MgCl₂ = 0.257 mols Mg(OH)₂ x 1 mol MgCl₂ / mol Mg(OH)₂ = 0.257 mols MgCl₂

mass MgCl₂ = 0.257 mols MgCl₂ x 95.2 g MgCl₂/ mol = 24.5 g MgCl₂