Question

3Cl2 + 6 NaOH → 5 NaCl + NaClO3 + 3H20 (Cl = 35.5, Na = 23)
How many gms of NaClO3 is produced from 2 mole chlorine gas when treated with excess
of NaOH.

​

Answer 1

Answer:

as every 3 moles of chlorine gas produces 1 mole of NaClCO3

so 2 moles of chlorine gas produces 2/3 moles  of NaClCO3

if the molar mass of NaClCO3= (23+35.5+3x16)=106.5

so the massof NaClO3 is produced from 2 mole chlorine gas when treated

with excess of NaOH= 2/3x 106.5=71 gram