Question

1. In the reaction Mg(OH)2 + 2HCl → MgCl2 + 2H2 O, you were given 15

grams of Mg(OH)2 and 25 grams of HCl. Which of the two reactants is the

limiting reagent? How much MgCl2 will be produced in the reaction? (Mg(OH)2

= 58.32 g/mol, HCl = 36.46 g/mol, MgCl2 = 95.32 g/mol)

Answer 1

From the equation 1 mole of Mg(OH)₂ reacts with 2 moles of HCl.

That means that $58.32 g$ of Mg(OH)₂ react with $2*36.46=72.92 g$ of HCl.

15 g of Mg(OH)₂ will need $15*72.92/58.32 = 18.76g$ of HCl. Since we have 25 g of HCl, Mg(OH)₂ is the limitting reagent.

Find the mass of MgCl₂:

$m(MgCl_2) = m(Mg(OH)_2)*M(MgCl_2)/M(Mg(OH)_2) = 15*95.32/58.32 = 24.52 (g)$

Answer: Mg(OH)₂ is the limitting reagent. 24.52 g of MgCl₂ will be produced in the reaction.