wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The reaction
Cl2+S2O32+OHSO42+Cl+H2O
Starting with 0.15 mole Cl2,0.010 moleS2O32and 0.30 mole OH, mole of Cl2 left in solution will be

A
0.11
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
0.01
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
0.04
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
0.09
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 0.11
4Cl2+S2O32+10OH2SO42+8Cl+5H2O
limiting reagent is S2O32
1 mole of S2O32 react with 4 mole Cl2
so,
0.01 mole S2O32=0.01×4=0.04moleCl2
moles of Cl2 left=0.150.04=0.11

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Concentration of Solutions
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon