Answer to Question #146121 in Physical Chemistry for Kaye Garvin

Question

If 16.0 g of Mg (OH)2 and 11.0 g of HCI are combined...what is the limiting reactant and how many grams of Mg CI2 will be produced?

Answer 1

The equation is

Mg(OH)₂ + 2HCl --> MgCl₂ + 2H₂O

$n_{Mg(OH)_2}=\frac{16.0g}{58.32g/mol}=0.274 mol$

$n_{HCl}=\frac{11.0g}{36.46g/mol}=0.302mol$

According to the equation, the mole ratio HCl : Mg(OH)₂ = 2 : 1

$\frac{0.302}{0.274}<\frac{2}{1}$ , hence HCl is the limiting reactant.

$m_{MgCl_2}=0.302mol_{HCl}\times\frac{1 mol_{MgCl_2}}{2mol_{HCl}}\times\frac{95.21g_{MgCl_2}}{1mol_{MgCl_2}}=14.4g$

Answer: HCl; 14.4 g