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2 years ago

How many moles of N2O5 are needed to produce 7.90 g of NO2 2N2O5 = 4NO2 + O2

2 answers:

8 0

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.

Furkat [3]2 years ago

7 0

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.

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