Question

How to calculate the weight percent Na2CO3 in impure base sample???

I did my calculations but it somehow doesnt make sense

First we find the moles of HCl

=0.1M HCl x volume of HCl used to titrate

       =0.1M x 10.5mL HCl

       =1.05 mol HCl

The ratio between Na2CO3 and Hcl is 1:2

Moles Na2CO3

= moles of HCl/2

=1.05mol HCl/2

=0.525mol Na2CO3

Mass Na2CO3

= moles of Na2CO3 x 105.9906 g/mol

       =0.525 mol Na2CO3 x 105.9906 g/mol

       =55.65 g

%    = mass of Na2CO3 x100/mass of sample from the impure solid

       =55.65g x 100/ 0.1328g = 41905.1205 ???

What did I do wrong??? (do I need to include the fact that that the impure base was first mixed with 25 mL of distilled water??? The HCl used is 0.1M for titrate to its endpoint)

Impure base	1
Mass (g)	0.1328
Trial 1 volume of HCl used in titration (mL)	10.5
Trial 2 /2nd titration(after boiling) : volume of HCl used in titration (mL)	3

Answer 1