How many moles of N2O5 are needed to produce 7.90 g of NO2

2N2O5 = 4NO2 + O2

Answer is: 9,27 g of nitrogen (V) oxide.

Chemical reaction: 2N₂O₅ → 4NO₂ + O₂.

m (NO₂) = 7,90 g.

n (NO₂) = m (NO₂) : M (NO₂)

n (NO₂) = 7,90 g : 46 g/mol

n (NO₂) = 0,17 mol.

from reaction n (N₂O₅) : n (NO₂) = 4 : 2

n (N₂O₅) : 0,17 mol = 2 : 1

n (N₂O₅) = 0,085 mol.

m (N₂O₅) = 0,085 mol · 108 g/mol.

m (N₂O₅) = 9,27 g.

n - amount of substance.

From the chemical eqn

No of moles of N2O5 = Mass / molar mass

Molar mass = (14 * 2) + (16 * 5) = 28 + 90 = 118g.

From the chemical equation 7.9g of NO2 reacts with 7.9 * 2 g of N2O5. We have a 1 : 2 mole ratio. Hence mass of N2O5 = 14.8g

No of moles = 14.8 / 118 = 0.133 moles.