Question

For 1 kg of CO, calculate the hydrogen required and the products produced for (a) ethylene (C2H4), (b) isopropanol (C3H80), and (c) pentane (C5H12). Water is a product for three cases. Note that the unknown coefficients in a chemical reaction can be obtained through mass balance written for each element. For example, you may write ethylene formula as CnH2n and find n.

Answer 1

solution 28 Mass of co m, = 1kg Molar mass of co - 28 kg Ikmal 8. Moles of co= ļ kmal = 0.03571 kmol B = 35.71 moles (a) Ethylene Production (Carta) co + H2 y Cahu + ZH₂O Now we make atomic species balance carbon balance cc) c in reactants = 1 moles c in products = ay reactants - Products oo l = 2y y = oxygen balance Co - o atom in left hand side (lints) atom o atom in Right hand side (R.His) Oo Now LiH.S = Rites . [Z=1 Hydrogen balance (H) 7 H in Litos = 2x OH in R.Hos = 4y+22

8 H in Lores a H in Rius 2x = 44+22 2x = 4x4 + 2x1 = 4 -) X=ų = 2 moles 8. Balanced reaction to produce Caly co + 2H₂ tahu + H2O 8 Moles of H₂ required for 35.71 moles of co A nna = 35.71X2 = 71.42 mdes & Mass of H₂ required for ethylene MH = 71.42x2 = 142.84 g Required mr2 = 0.14284 kg ethylene molar mass is 28 kg 1 kmol Moles of ethylene produece = 2 x 35.71 moles = 17,855 moles - mass of Cama Produce mohu = 17.855 x 28 gramy - mot = 499.94 gram = 0.5 kg u y modes of H₂O Produece = 35.71 moles 8. mass of H₂O Produce = 35.71X18 g Im H2O = 642-739 [MH2O = 0.64272 kg

(6) Isopropanol, CC3H80) Reaction cotx H₂O Y G Hg O + 2H₂O - c- balance l o-balance H-balance 3y = 1 ly= 201= y +2 | 2x = 8y +22 Hei-*= 21 22 = 8X1 + 2x2 2x = 4 [2 = 3/3 12=2] 80 Balance reaction - Cot 2 H₂ → 1 C3HgO + 2 H2O si 8 Mass of H₂ required = (2x 35.71) x2 g required I m2 = 0.1428 kg 4 Moles of C3HgO Produece = + x 35.71 moles = 11.9033 moles molar mass of C3H80 = 60 g/mole • Mass of CzHg O Rroduece = 60x11.903? meg Hgo = 714.7 98 8 mezuço = 0.11197) U Moles of H₂O Produca = 2 x 35.41 = 23.806 moles o. Mass of H₂o Boduece = 23.806 X 18 g muo 428.929 Imago =0,428528

(②) Pentane (C5H12) Reaction CotxH₂ = USHI2 + 2H2O C-balana o-balance I H-balanca 5y = 1 Iz [yo5 | Zall 2x = 12y + 22 2x = 19XL + 2 2x + 23 22 = 12 + 10 TO 2x = 22 + x=32 2 = 2-2) 8. Balanced reaction Cot 2:2 H₂ + H2+ ho %. Mass of Hy required = (2-2835.71) X2 g IMH2 = 0.157124) kg 4 Molar mass Pentane = 72 g/mol Moles of Pentane Produece = tx 35.71 = 7,142 moles & Mass of Pentane required motio 7.142872 g I mashi2 = 514.224 g = 0.5142 kg/ ♡ Matatags moles of H2O Produece = 1X35.71= 35.71 • Mass of H₂O Produece = 35.71x 18 g moles I mH₂O = 642.788 = 0.6427 kgl.