Answer to Question #349100 in General Chemistry for Chemistry

Question

A 1.68 gram sample of CaCO₃ (100.09 g/mol) is placed in a 109 mL flask and dissolved in HCl, calculate the resulting molarity of the calcium ion.

Answer 1

m(CaCO₃) = 1,68 g

M(CaCO₃) = 100,09 g/mol

V = 109 ml = 0,109 l

C_M = ?

1.

$n= \cfrac {m}{M}= \cfrac {1,68g}{100,09g/mol}=0,0168mol$

CaCO₃ + 2HCl = CaCl₂ + CO₂ + H₂O

1mol CaCO3 => 1mol Ca²⁺

0,0168mol CaCO₃ => X

X = 0,0168 mol Ca²⁺

2.

We find concentration of molarity of Ca²⁺

$C_M= \cfrac {n}{V}= \cfrac {0,0168mol}{0,109l}= 0,154mol/l$

Answer: C_M (Ca²⁺) = 0,154 mol/l