Ethanol at 25 C and 24.0% excess air at 25 C enter a reactor. The ethanol combusts completely, and the product gas exits the reactor at 120.0 C. The cooling coils limit the throughput of the reactor. If the cooling coils remove -3825 kJ/s, what is the maximum amount of ethanol that can be combusted in the system? You will answer a series of questions to arrive at this answer. 25 C, 1 atm ni mol C,H,OH/s 25 C, 1 atm n2 mol O2/s n3 mol N / ooo 120.0 C, 1 atm n4 mol O2/S ng mol N / ng mol CO2/s n7 mol H, Ols O = -3825 kJ/S The reaction is C,H,OH(1) + 302(g) 2002(g) + 3 H, 0(1) A = -1366.9 kJ/mol First, express the molar flow rates of O, and N, entering the reactor and the molar flow rates of all product gases leaving the reactor in terms of n, the molar flow rate of ethanol into the reactor. For example, if there was an ng and it equaled n, you would put 1.5 in the answer field. If ng = n, then put 1 in the answer field. mol O. n2 = .ni S mol N n3 = ni S mol O n4 = .ni S mol N ni ng = mol CO2 n6 = ni mol H, O n7 = ni S Take ethanol, O2, and N, at 25 C to be the reference states when finding the given specific enthalpies. What is 4, the specific enthalpy of O, at 120.0 C? 4 = kJ/mol What is 5, the specific enthalpy of N, at 120.0 C? n = kJ/mol What is 6, the specific enthalpy of Co, at 120.0 C? 16 = kJ/mol What is 1, the specific enthalpy of steam at 120.0 C? , = kJ/mol What is n, the maximum molar flow rate of ethanol entering the reactor? ni = mol C,H,OH/s