a) What theoretical mass of O2 will be needed to burn 36.1 g of B2H6? (b) What mass of O2 will be needed to burn 36.1 g of B Ho if O2 is 60% in excess? (c) What volume of 60% in excess air at 1 atm and 25 deg C is needed to burn 36.1 g of B2H? (d) How many moles of water are produced from 36.1 g of B2H6?

One kmol of benzene gas (C6H6) enters a combustion chamber designed to operate at steady state where it is mixed and burned with air. C,Hạ + a (O, +3.76N,)→ BCO, + yH,O+8N, Determine the following: 54. the balanced reaction equation for the theoretical amount of air, if the combustion products include CO2, H20 and N2, only. 55. the balanced reaction equation for 30 percent excess air, if all the hydrogen in the fuel burns to H20, but only 92 percent of the carbon burns to CO2, while the remaining 8 percent burns to CO. There is also some free Oz in the product gases. 56. the air-fuel ratio on a molar basis, in kmol air / kmol fuel (for 30 percent excess air); 57. the air-fuel ratio on a mass basis, in kg air / kg fuel (for 30 percent excess air); 58. the mass flow rate of the air entering the chamber, in kg air / min, if the fuel enters the chamber at a rate of 0.05 kg fuel / min (for 30 percent excess air); 59. the dew point temperature of the combustion products, in °C, when cooled…

5. Hydrogen H2 is burned completely with the stoichiometric amount of air during a steady state, stead flow combustion process. If both reactants and products are maintained at 25 C and 1 atm, and the water in the products exists in the liquid form, determine the heat transfer from the combustion chamber during the process. What would your answer be if the combustion were achieved with 80% excess air ? Ans. (a) Q = 285,830 ku/kmol H2, (b) challenge *6 at 25 L

One strategy for dealing with the acidification of lakes is periodically to add powered limestone (CaCO3) to them, resulting in pH of 10.95. Calculate the Ca2+ concentration in the lake. CO2 in atmosphere 370 ppm KH,CO2 = 0.033 mol/(L∙atm) Ka1 = 10-6.3 Ka2 = 10-10.3 Ksp,CaCO3 = 4.57 x 10-9