Question

Estimate the theoretical amount of gas (ft3 of methane and carbon dioxide) in that can be produced under anaerobic conditions from one pound of sucrose(C12 H22 O11) at STP conditions.

Answer 1

The balanced chemical equation is given as

C₁₂H₂₂O₁₁ (s) + H₂O (l) ---------> 6 CH₄ (g) + 6 CO₂ (g)

As per the stoichiometric equation,

1 mole C₁₂H₂₂O₁₁ = 6 moles CH₄ = 6 moles CO₂.

Therefore, 1 mole of sucrose furnishes (6 +6) moles = 12 moles of gaseous products.

The atomic masses are

C: 12.011 u

H: 1.008 u

O: 15.999 u

The gram molar mass of sucrose = (12*12.011 + 22*1.008 + 11*15.999) g/mol = 342.297 g/mol.

It is known that 1 pound = 453.592 g.

Therefore, mole(s) of sucrose corresponding to 453.592 g = (mass in grams)/(molar mass in grams/mole)

= (453.592 g)/(342.297 g/mol)

= 1.325 mole.

As per the stoichiometric equation,

moles of gases produced = (1.325 mole sucrose)*(12 moles gases)/(1 mole sucrose)

= 15.9 moles gases.

It is known that

1 mole of a gas occupies 22.4 L at STP.

Therefore, the volume occupied by 15.9 moles of gases at STP = (15.9 moles)*(22.4 L)/(1 mole)

= 356.16 L

= (356.16 L)*(1 ft³)/(28.3168 L) [1 ft³ = 28.3168 L]

= 12.5777 ft³

≈ 12.6 ft³ (ans).