Question

If 16.0 g of Mg (OH)2 and 11.0 g of HCI are combined...what is the excess reactant?

Answer 1

Mg(OH)₂ + 2HCl = MgCl₂ + 2H₂O

For the reaction with 1 mol Mg(OH)₂ 0.5 mol HCl is needed

n = $\frac{m}{M}$

n(Mg(OH)₂) = $\frac{16.0}{58,4} = 0.274 (mol)$

n(HCl) = $\frac{11}{36.5} = 0.31 (mol)$

$\frac{0.31}{2} = 0.165$

Answer: Mg(OH)₂ is an excess reactant.