300 L/h of 20% mol (30% = C₃H₈, 80% n-C₄H₁₀ mixture at 0°C and 1.1 atm and 500 L/h of a 45% mol (45% = C₃H₈, 55% n-C₄H₁₀ mixture at 25°C and 1.1 atm are mixed and heated to 227°C at constant pressure.Enthalpies of propane and n-butane are listed below. Assume ideal behavior.| °C | Propane ΔH(J/mol) | Butane ΔH(J/mol) ||-----|-------------------|------------------|| 0 | 0 | 0 || 25 | 1772 | 2394 || 227 | 20,965 | 27,412 |Calculate the heat requirement in kJ/h.

Given the system consists of two inlets and one outlet. Inlet(1): Flow rate of gas, F1 = 300 L/h mole fraction of propane, x1 = 0.20 Temperature, T1 = 0 °C = 273 K Pressure, P1 = 1.1 atm Since the gas is ideal gas, at STP conditions T = 273 K, P = 1 atm, 1 mole of gas occupies 22.4 L of volume. Calculate the molar flow rate of the gas as follows: n1 = F1×P1P×1 mol22.4 Ln1 = 300 Lh×1.1 atm1 atm×1 mol22.4 Ln1 = 14.73 mol/hInlet(2): Flow rate of gas, F2 = 500 L/h mole fraction of propane, x2 = 0.45 Temperature, T2 = 25 °C = 298 K Pressure, P2 = 1.1 atm Since the gas is ideal gas, at STP conditionsT = 273 K, P = 1 atm, 1 mole of gas occupies 22.4 L of volume. Calculate the molar flow rate of the gas as follows: n2 = F2×P2P×T2T×1 mol22.4 Ln2 = 500 Lh×1.1 atm1 atm×298 K273 K×1 mol22.4 Ln2 = 26.8 mol/hCalculate the number of moles of propane at outlet as follows: nC3H8 = n1x1 + n2x2 = 14.73 mol/h×0.20 +26.8 mol/h×0.45 =2.95 +12.06 mol/h =15.01 mol/h Calculate the number of moles of butane at outlet as follows: nC4H10 = n11-x1 + n21-x2 = 14.73 mol/h×0.80 +26.8 mol/h×0.55 =11.784 +14.74 mol/h =26.52 mol/h Write the energy balance as follows: Q = enthalpy at outlet -enthalpy at inletsQ = nC3H8 ×H⏞C3H8 at 227 °C+ nC4H10 ×H⏞C4H10 at 227 °C -n2x2 ×H⏞C3H8 at 25 °C+ n2(1-x2) ×H⏞C4H10 at 25 °CQ=15.01molh×20685 Jmol+26.52molh×27442 Jmol-26.8 molh×0.45×1772 Jmol +26.8molh×0.55×2394 JmolQ = 310481.85 + 727761.84-21370.32 - 35287.56Jh×1 kJ1000 J Q = 981.6 1 kJhTherefore, the required solution is 981.6 1 kJh.