Question

12. Propane (C3H8) is used as a fuel in many gas grills. The balanced equation for the combustion of C3Hg is shown in Equation 10. C3H8(g) +502(g) + 3 CO2(g) + 4H2O(g) + energy (Eq. 10) (a) How many grams of CO2 could possibly be produced if 10.0 g of CzHg reacts with an ample supply of Oz? (b) How many grams of CO2 could possibly be produced with an ample supply of C3Hg, but only 10.0 g of O2? (c) If this reaction is run using 10.0 g of C3Hg and 10.0 g of O2, what mass of CO2 could possibly be produced? (d) Identify the limiting reagent in the reaction described in (c). mida (R .) reacts with H2O to form H3PO4. y

Answer 1

CH&G + son 3 Corg, +4 Hogy + Energy Imole smoles 3 moles 4 moles Given Cg Hg = 10oog Moles of Cate = 10.off =0.2268 mol mob Line Enadarmoss who motoren = 44•19 mai of CHE = 44• Iglmal I mole Cattg gives 3 moles co2 0.2268 mole (Hg gives a moles con ': x= 3x0.2268 moes com <= 0.6804 moles coz 0.6884 moles coa 0.6884 mol +44001g fimplar mass of 244.01g/ml = 29.94 g Cliven O₂ = 10:0g Modes of g - toode 0.3125 ml E made ones = 31.-99 35 hed] mol L co₂ w - smole o gives 3 moles cor 0.3125 mle og gives x'ndes co x= 3x0.3125 moles con . x= 0.1875 molesco, 20-1875 moles co2 = 0.1875 ml x 44001g = 8.2529? wote

© Given log Hg = 0.2268 modes Cats log a = 0.3125 moles o I mole Caitg requines smoles o 0.2268 mdles Stg requines 'x'ndes a x= 0·2268 moley cits * Smoles on moleshe [x = 11134 moles of Given that reaction is ren using 1010g of o, which means 0.3125 roles of In the limiting agent is a l Hence, Calculate the mass of co produced by using the number of moles of limiting reagent of s moles 2 = 3 moles co₂ 0.3125 moles ₂ = sc molesco, x= 3 moles com x0.3125 molesa s moles or x= 0.1875 moles com 0-1875 moles co, I 01875 mol x 44.012 = 8.25 g co g produced, The limiting breagent is 2. The excess reagent & Catg @