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.

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