Answer:
There are 6 mol of NO2 with respect to 3 mol of N2O5
Step-by-step explanation:
Approach 1 ( dimensional analysis ) :
3 Moles of N2O5
( 4 moles of NO2 / 2 Moles of N2O5 ) - moles of N2O5 cancel out, leaving you with the moles of NO2 -
3
4 / 2 = 12 / 2 = 6 moles of NO2
So as you can see in the formula there are 4 moles of NO2 present per 2 Moles of N2O5 - " 4NO2 and 2N2O5. " As we wanted the moles of N2O5 to cancel out, the 2 moles of N2O5 was kept as the denominator, and hence we received the fraction we needed.
Approach 2 :
There are 3 Moles of N2O5. The ratio of Moles of N2O5 to moles of NO2 is provided by the reaction -
Moles of N2O5 : Moles of NO2,
2 : 4,
1 : 2
Therefore the moles of NO2 will be two times as much as the given moles of N2O5, or 3
2 = 6 moles of NO2