Question

448 ml of So2 at NTP is passed through 100ml of 0.2N solution of NaOH. Find the weight of the salt formed​

Answer 1

Answer:

Explanation:

weight of salt formed = limiting ratio*coefficient

0.02 mole so2

moles*f*1000=20

moles=0.002

0.002 mole naoh

make oxy acid of so2 which is h2so3

make 1 degree salt of that

nahso3 will be formed

so2+naoh=nahso3

limiting ratio which is 0.002

weight of salt formed = 0.002*104

=2.08 gm