Question

7. A. Balance the following combustion reaction of pentane(CsH12): (16 pts) CsH12 + O2(g) + CO2(g) + H2O(9) B. If I begin with 6.85 grams of pentane how many grams of CO2 will theoretically be produced? C. What would the temperature(K) of CO2 gas be if the amount produced in Part B had a pressure of 1.55 atm. and occupied 27.3 liters?

Answer 1

The geven Combustion seaction is Cs Hip + 0219) >20217) + H20(3) and its Balanced equation isnaby Csie + 80,(9) - 5CO2614 6H. 6H2005) No. of moles of pentane = geven mass of est2 Molar mass of C5H12 6.85 5x12+ 12x1 mol Sob 221 6.85 +2 mol. 1 mol of estic produces s moles of CO2 (the produce sx .6.85 mol of CO2 6.85 mol of 72 T 72 T - 34.25 mol of CO₂ 9.72 ula n = 0.475 69 mol of CO2 No. of moles of CO2 0.47569 mol given mass of CO2 0.47569 Molar mass of CO2 = 0.47569 & Gwen mass of CO2 12+ 32 se geven mass of CO2 = 20.93 gram

CCJ Por CO2 Rz0-0821 Latm ktmolt 2 Pressure; B = 1. ss alm Volume, Vis 27.3 Luna no, of moles, 13 0.47,569 moles We know that wahre PV z NRT T2 DV ten NR T. T is 1:55 27.3 0-47569% 0.0821 so ri 2012 predam 42.315 → DE TIF 0.0391 T 1.082.23 ki Temperature of CO2 gas is 1082.23k. sobre o corpo e