Question

1. The ultimate analysis shows that the C, H, O, N and S contents of a biomass material are 51.9%, 5.5%, 41.5%, 0.8% and 0.3% on a dry basis. What is the chemical formula of this biomass? How many kilograms of air are required to completely combust 1 kg of this biomass?

Answer 1

Basis - 100 g of biomass

Moles of elements = mass/molecular weight

Moles of C = 51.9/12.011 = 4.32

Moles of H = 5.5/1.0079 = 5.46

Moles of O = 41.5/15.9994 = 2.59

Moles of N = 0.8/14.0067 = 0.06

Moles of S = 0.3/32.06 = 0.01

Mol ratio

C : H : O : N : S = 4.32 : 5.46 : 2.59 : 0.06 : 0.01

Divide by 0.01

Mol ratio

C : H : O : N : S = 432 : 546 : 259 : 6 : 1

Chemical formula = C₄₃₂H₅₄₆O₂₅₉N₆S

Moles of biomass = 1000g /9999.0102 g/mol

= 0.100 mol

The balanced combustion reaction of biomass

C₄₃₂H₅₄₆O₂₅₉N₆S + 440O₂ = 432CO₂ + 273H₂O + 3N₂ + SO₂

1 mol biomass required = 440 mol O2

0.100 mol biomass requires = 440 mol O2 * 0.100

= 4.4 mol O2

1 mol air consists 0.21 mol O2

Moles of air required = 4.4/0.21 = 20.95 mol air

Mass of air required = 20.95 mol x 28.97 g/mol x 1kg/1000g

= 0.607 kg