Question

What mass of MgCl2 was produced from 5.87 of Mg(OH)2 reacting with 12.84 g of HCl? Identify the limiting and the excess reactants.

Answer 1

Answers;

Mass of MgCl2 = 9.615 g

Mg(OH)₂ i limiting reactant

HCl is the excess reactant

Explanation;

The equation for the reaction is;

Mg(OH)2+ 2 HCl → MgCl2+ H2O

5.87 g Mg(OH)2x(1mol Mg(OH)2/58.31g) = 0.101 moles OF Mg(OH)2

Moles of MgCl2 = 0.101 moles

Mass of MgCl2 = 0.101 × 95.2g/mole

= 9.615 g MgCl2

Moles of HCl

0.101 x (2 mol HCl/ 1 mol Mg(OH)2) = 0.202 moles

Mass of HCl

0.202 moles x(36.45 g HCl/1 mol HCl)=7.34 g

Therefore; 7.34 g of HCl are needed out of 12.84 g HCl given,

Therefore; Mg(OH)2 is the limiting reactant and HCl is the excess reactant