this is a Chemical Engineering question. Problem 4: A mixture of propane and hyd
ID: 938485 • Letter: T
Question
this is a Chemical Engineering question.
Problem 4: A mixture of propane and hydrogen is burned with excess air. The exiting exhaust gases contained 11.8 mol H20. After all of the water was removed, the residual gas contained (in mol%): 0.34% C3HS: 0.17% H2, 8.84% CO2: 0.46% CO: 6.49% O2 and 83.69% N. a) Taking a basis of 100 mol dry gas, leaving the furnace, determine molar amounts of propane, hydrogen, carbon dioxide, carbon monoxide, nitrogen, oxygen and water b) Write balanced equations for all reactions that occur in the furnace c) Determine the amount of O2 fed to the reactor and the %XS hint: N2 is inert d) Determine the ratio of mol C3Hsmol H in the feed gas (suggestion: use atomic balances.Explanation / Answer
(a) Given the exiting exhaust gas contains 11.8 mol% H2O.
Hence the percentage of exiting exhaust gas that is dry air = 100 % - 11.8% = 88.2 %
Given 88.2% of exiting exhaust gas is equal to 100 mol.
Hence moles of water in the exiting exhaust gas = (100 mol / 88.2) x 11.8 = 13.4 mol H2O (answer)
moles of propane = 100 mol x (0.34 / 100) = 0.34 mol (answer)
moles of hydrogen(H2) = 100 mol x (0.17 / 100) = 0.17 mol (answer)
moles of carbon dioxide = 100 mol x (8.84 / 100) = 8.84 mol (answer)
moles of carbon monoxide = 100 mol x (0.46 / 100) = 0.46 mol (answer)
moles of oxygen = 100 mol x (6.49 / 100) = 6.49 mol (answer)
moles of nitrogen = 100 mol x (83.69 / 100) = 88.69 mol (answer)
(b) Balanced chemical reaction is
3.10 C3H8 + H2 + 15.77 O2 ------ > 8.84 CO2 + 0.46 CO + 13.4 H2O
(c) Amout of O2 feed to the reactor = 8.84 mol in CO2 + 0.23 mol in CO + 6.70 mol in H2O + 6.49 mol in exhaust air = 22.26 mol O2. (answer)
(d) Moles of C3H8 in the feed gas = 3.10 mol + 0.34 mol = 3.44 mol
Moles of H2 in the feed gas = 1 + 0.17 = 1.17 mol
Hence ratio of C3H8 to H2 = 3.44 / 1.17 = 2.94 (answer)
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