Stoichiometry

100 ml of 0.1 M NaAl(OH)2CO3 is neutralised by 0.25 N Hcl to form Nacl ,Alcl3 & CO2

Volume of Hcl reqd is ...

(A) 10 mL

(B) 40 mL

(C) 100 mL

(D) 160 mL

9 Answers

19
Debotosh.. ·

(100 * 0.1) = ( 0.25 * v)/2
=>20/0.25 =v =80ml

24
eureka123 ·

the main prob i suppose will be in judging eq wt of NaAl(OH)2CO3
which is 1
becoz Na+ and (Al3+(OH-)2CO32-)-

*edit :its not eq wt as i wrote...its n factor *

19
Debotosh.. ·

if in a balanced chemical equation , we find x* acid + y* base gives the correct balance, then
V1M1 / X = V2M2 / Y
(M-MOLARITY )

4
UTTARA ·

Thanks

1
decoder ·

answer should be 160ml naa!!!!

n-factor of given compound is 4
so by eq.concept
100*0.1*4=0.25*V
V=160ml

4
UTTARA ·

No decoder I guess n - factor is 1

Check out

as eureka said Na+ & ...

19
Debotosh.. ·

what is the answer ?

4
UTTARA ·

Answer is 80 mL

19
Debotosh.. ·

so thats the process !

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