CALCULATOR FULL SCREEN PRI Problem 4.84 Combustion Mixture A gas contains 65.0 w

Question

CALCULATOR FULL SCREEN PRI Problem 4.84 Combustion Mixture A gas contains 65.0 wt% methane, 5.00 wt% ethane, 20.0 wt% ethylene, and the balance water. Composition Correct. Calculate the molar composition of this gas on both a wet and a dry basis and the ratio (mol H20/mol dry gas) Component Mole % (wet basis) Mole % (Dry basis) Methane 73.877 82.179 3.371 hane 3.031 Ethylene 12.99 10.102 14.450 Water LO mol H20/mol dry gas: 0.1123 Open blank Show Work Copy Show Work from Previous attempt Show Work is REQUIRED for this question:

Explanation / Answer

4.84

Take basis = 100g fuel

molar mass of methane = 16g/mol , mass of methane = 65g

moles of methane = 65g/ 16g/mol = 4.0625 mol

molar mass of ethane = 30g/mol , mass of ethane = 5g

moles of ethane = 5g/30g/mol = 0.16667 mol

molar mass of ethylene = 28g/mol , mass of ethylene = 20g

moles of ethylene = 20g/28g/mol = 0.7143 mol

molar mass of water = 18g/mol , mass of water = 10g

moles of water = 10g/18g/mol = 0.5555 mol

total moles = 0.5555 mol + 0.7143 mol + 0.16667 mol + 4.0625 mol = 5.50 mol

mole % of methane = 4.0625 / 5.50 x 100 = 73.86

ethane = 3.03

ethylene = 12.99

water = 10.10

On dry basis(without water) total moles =  5.50 mol - 0.55555 mol (water) = 4.94

methane moles % = 4.0625 / 4.94 x 100 = 82.24

ethane = 3.37

ethylene = 14.46

moles of H2O / mol dry gas = 0.5555 mol / 4.94 mol = 0.1124

For 100g we already have no. of moles of each component

Balanced chemical equation for combustion are:

CH4 + 2O2 -----> CO2 + 2H2O

4.0625 mol methane will need = 2 /1 x 4.0625 = 8.1250 mol O2

2C2H6 + 7O2 ----->4CO2 + 6H2O

0.166667 mol ethane will need = 7/2 x 0.16666667 = 0.5833 mol O2

C2H4 + 3O2 ------->2CO2 + 2H2O

0.7143 mol ethylene will need = 3/1 x 0.7143 = 2.1429 mol O2

total moles of O2 needed per 100g fuel = 2.1429 mol + 0.5833 mol +  8.1250 mol = 10.8512 mol

21 mol O2 is present in 100 mol air

for 10.8512 mol O2 we need = 100/21 x 10.8512 = 51.6724 mol air

30 % extra = 51.6724 mol + 51.6724 mol x 30/100 = 67.17 mol air per 100g feed

for 100 kg feed = 67.17 kmol air/h

For 75% conversion , air needed = 51.6724 mol x 75/100 = 38.7543 mol/h

for 100 kg/h = 38.7543 kmol/h

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