Calculate how many moles of NO2 form from 3.0 mol N2O5 2N2O5(g)→4NO2(g)+O2(g)Express your answer using two significant figures.

There are 6 mol of NO2 with respect to 3 mol of N2O5

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

b) nuclear ive learned that in a pedtrodic table