Question

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.

Answer 1

Answer:

6.0 moles NO2(g)

Explanation:

Based on the reaction every 2 moles N2O5(g) gives reaction with 4 moles NO2(g).Then when we have 3.0 mol N2O5(g),

2 moles N2O5(g)       4 moles NO2(g)

3 moles N2O5(g)        ? moles NO2(g)

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3 *4 / 2 = 6.00 moles NO2(g)

Answer 2

Answer:

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