Question

Assume the reaction of 27.58 g of O₂ in the following reaction to determine the other quantities.

7O2 + 4Mn -> 2Mn2O7
moles of O2 reacting =	mol
moles of Mn required =	mol
moles of Mn2O7 formed =	mol
mass of Mn2O7 formed =	g

Answer 1

1)

Molar mass of O2 = 32 g/mol

mass of O2 = 27.58 g

we have below equation to be used:

number of mol of O2,

n = mass of O2/molar mass of O2

=(27.58 g)/(32 g/mol)

= 0.8619 mol

2)

from reaction,

moles of Mn reacting = (4/7)*moles of O2 reacting

= (4/7)* 0.8619 mol

= 0.4925 mol

3)

from reaction,

moles of Mn2O7 reacting = (2/7)*moles of O2 reacting

= (2/7)* 0.8619 mol

= 0.2463 mol

4)

Molar mass of Mn2O7 = 2*MM(Mn) + 7*MM(O)

= 2*54.94 + 7*16.0

= 221.88 g/mol

we have below equation to be used:

mass of Mn2O7,

m = number of mol * molar mass

= 0.2463 mol * 221.88 g/mol

= 54.65 g

Answer: 54.65 g