One kilogram of "as received" yard trimmings is made up of approximately 620 g moisture, 330 g of decomposable organic matter (represented by molar composition C1276H21 280 26Nosa) and 50 g of other non-biodegradable constituents. Calculate: (a) the moles of methane gas produced from one mole of decomposable organic matter, (b) the volume of methane gas (0.224 m /mole) from 1 kg of "as received" yard trimmings, and (c) energy (890 kJ/mole CH that could be produced from anaerobic decomposition of one kilogram of "as received" yard trimmings in a landfill. [Given: C, HONn H0 m CHs CO2 + d NH and m (4a + b-2c-3d)/8]

Explanation:

(a) Given that 620g moisture and 330g decomposable organic matter in yard trimming is represented by C₁₂.₇₆H₂₁.₂₈O₉.₂₆N₀.₅₄

Given the atomic mass of Carbon C = 12, Hydrogen H = 1, Oxygen O = 16 and Nitrogen N = 14

1 mole of trimming = 12*12.76 + 1*21.28 + 16*9.26 + 14*0.54

=  153.12 + 21.28 + 148.16 + 7.56

= 330.12 g/mol

which means 1 kg of as received trimming has 330 g of decomposable that produce 1 mole of decomposable

The moles of methane produced will be given as

m = (4a + b -2c - 3d)/8

= (4*12.76 + 21.28 - 2*9.26 - 3*0.54)/8

= (51.04 + 21.28 - 18.52 - 1.62)/8

= 52.18/8

= 6.5225

(b) Volume of methane V is given as

V = (0.0224 m³ CH₄mol/CH₄) × (6.5225 mol CH₄/ kg)

= 0.1461 m³ CH₄/kg lawn trimmings

(c) Energy will be given as

CH₄Energy = 6.5225 mol of CH₄/kg × 890 kJ/mol

= 5805.025

≈ 5805 kJ/kg

answer:

yes

explanation:

answer:

the answer is d

explanation: