Question

1. The neutralization reaction between HCl and Ca(OH)2 produces CaCl2 and water. If 0.75g of Ca(OH)2 is used in the reaction, determine the following:
a) Number of moles of HCl needed to completely react with Ca(OH)2¬
b) Mass of CaCl2 formed in the reaction.

Answer 1

Solution.

$m(Ca(OH)_2)=0.75g;$

$2HCl+Ca(OH)_2 \to CaCl_2+2H_2O;$

$a) M(Ca(OH)_2)=74g/mol;$

$\nu=\dfrac{m}{M}; \nu(Ca(OH)_2)=\dfrac{0.75g}{74g/mol}=0.01mol;$

$x mol$ $0.01mol$

$2HCl+Ca(OH)_2 \to CaCl_2+2H_2O;$

$2$ $1$

$x=\dfrac{2\sdot 0.01mol}{1}=0.02mol;$

$\nu(HCl)=0.02mol;$

$b) M(CaCl_2)=111g/mol;$

$0.75g$ $xg$

$2HCl+Ca(OH)_2 \to CaCl_2+2H_2O;$

$74g/mol$ $111g/mol$

$x= \dfrac{0.75g\sdot111g/mol}{74g/mol}=1.125g;$

$m(CaCl_2)=1.125g;$

Answer: $a) \nu(HCl)=0.02mol;$

$b)m(CaCl_2)=1.125g.$