Calculate how many moles of NO2 form when each quantity of r

Calculate how many moles of NO2 form when each quantity of reactant completely reacts.
2N2O5(g)4NO2(g)+O2(g)

a. 3.0 mol N2O5

b. 7.2 mol N2O5

c. 17.0 g N2O5

d. 3.39 kg N2O5

Solution

from reaction,
moles of NO2 formed = (4/2)*number of moles of N2O5 reacting
= 2*number of moles of N2O5 reacting

a)
moles of NO2 formed = 2*number of moles of N2O5 reacting
= 2*3.0 mol
= 6.0 mol
Answer: 6.0 mol

b)
moles of NO2 formed = 2*number of moles of N2O5 reacting
= 2*7.2 mol
= 14.4 mol
Answer: 14.4 mol

c)

Molar mass of N2O5,
MM = 2*MM(N) + 5*MM(O)
= 2*14.01 + 5*16.0
= 108.02 g/mol


mass(N2O5)= 17.0 g

use:
number of mol of N2O5,
n = mass of N2O5/molar mass of N2O5
=(17.0 g)/(108.02 g/mol)
= 0.157 mol

moles of NO2 formed = 2*number of moles of N2O5 reacting
= 2* 0.157 mol
= 0.314 mol
Answer: 0.314 mol

d)

mass(N2O5)= 3.39 Kg
= 3390 g

use:
number of mol of N2O5,
n = mass of N2O5/molar mass of N2O5
=(3390.0 g)/(108.02 g/mol)
= 31.4 mol

moles of NO2 formed = 2*number of moles of N2O5 reacting
= 2* 31.4 mol
= 62.8 mol
Answer: 62.8 mol