For the reaction shown, calculate how many moles of \(\mathrm{NO}_{2}\) form when each amount of reactant completely reacts.

\(2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \rightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\)

(a) 1.3 mol \(\mathrm{N}_{2} \mathrm{O}_{5}\)

(b) 5.8 mol \(\mathrm{N}_{2} \mathrm{O}_{5}\)

(c) \(4.45 \times 10^{3} \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{5}\)

(d) \(1.006 \times 10^{-3} \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{5}\)

Equation Transcription:

Text Transcription:

NO_2

2 N_2 O_5 (g) right arrow 4 NO_2 (g) + O_2 (g)

N_2 O_5

4.45 times 10^3 mol N_2 O_5

1.006 times 10^-3 mol N_2 O_5

Solution 17P

Here, we are going to calculate the number of moles of NO2 formed when the given amounts of N2O5 completely reacts.

Step 1:

The given chemical reaction is:

2N2O5(g) → 4NO2(g) + O2(g)

From the above equation, it is clear that, 2 moles of N2O5 reacts completely to give 4 moles of NO2 as product.

Therefore, 1 mole of N2O5 will give (4/2 = 2) mol of NO2 on complete reaction.