Answer to Question #320166 in Physical Chemistry for Angela Caliso

Question

Directions: Show your complete solutions legibly for every item

1. Given the following reaction:

Mg(OH)2+ 2 HCI - MgCl2+ 2 H20

If 14.5.0 g of Mg(OH)2 and 12.0 g of HCI are combined

a. What is the limiting reactant?

b. How many grams of Mg Cl2 will be produced?

Answer 1

M(Mg(OH)₂) = 58 g/mol;

m(Mg(OH)₂) = 14.5 g;

M(HCl) = 36.5 g/mol;

m(HCl) = 12.0 g;

M(MgCl₂) = 95 g/mol;

n(Mg(OH)₂) = m(Mg(OH)₂)/M(Mg(OH)₂) = 14.5/58 = 0.250 mol;

n(HCl) = m(HCl)/M(HCl) = 12.0/36.5 = 0.329 mol;

Mg(OH)₂+ 2HCI = MgCl₂+ 2H₂O

By the chemical reaction:

n(Mg(OH)₂)' = n(Mg(OH)₂) = 0.250 mol;

n(HCl)' = 1/2 * n(HCl) = 1/2 * 0.329 = 0.164 mol.

So, HCl is the limiting reactant and Mg(OH)₂ is in excess.

n(MgCl₂) = 1/2 * n(HCl) = 1/2 * 0.329 = 0.164 mol.

m(MgCl₂) = n(MgCl₂) * M(MgCl₂) = 0.164 * 95 = 15.6 g.

Answer: HCl is the limiting reactant; m(MgCl₂) = 15.6 g.