Hello! I am given these questions as practical questions so I havetried to answe

Question

Hello! I am given these questions as practical questions so I havetried to answer but got stuck,please help (1)A 3.50L container of air at 670 torr and 360K has2.40cm3(cubic) of dichloromethane CH2Cl2 added,whichvaporizes.Calculate the total pressure, and the mole fractionof dichloromethane.[desity of dichloromethane is 1.325g/cm3] (2)The steel bomb is a bomb calorimeter,which has a volume of0.794L, is charged with Oxygen to a pressure of 11.0 atm at 27degrees Celcius. Calculate the moles of Oxygen in the bomb. 8.35gof solid Oxalic acid,(COOH)2, is burnt in the bomb to give carbondioxide and water .Assuming the gases behave ideally, what will thepressure in the bomb when the temperature returns to27degrees Celcius? Hello! I am given these questions as practical questions so I havetried to answer but got stuck,please help (1)A 3.50L container of air at 670 torr and 360K has2.40cm3(cubic) of dichloromethane CH2Cl2 added,whichvaporizes.Calculate the total pressure, and the mole fractionof dichloromethane.[desity of dichloromethane is 1.325g/cm3] (2)The steel bomb is a bomb calorimeter,which has a volume of0.794L, is charged with Oxygen to a pressure of 11.0 atm at 27degrees Celcius. Calculate the moles of Oxygen in the bomb. 8.35gof solid Oxalic acid,(COOH)2, is burnt in the bomb to give carbondioxide and water .Assuming the gases behave ideally, what will thepressure in the bomb when the temperature returns to27degrees Celcius?

Explanation / Answer

Here are my answers: Question 1: mass CH2Cl2 = density x volume =(1.325)(2.40) = 3.18 g CH2Cl2 Molar mass CH2Cl2 = 12.01 + 2(1.008) +2(35.45) = 84.93 g/mol CH2Cl2 => nCH2Cl2 = 3.18 / 84.93 = 0.0374 molCH2Cl2 PCH2Cl2 = nRT/V = (0.0374)(0.0821)(360)/(3.50) = 0.316atm = 240 torr => Ptotal = Pair + PCH2Cl2 =670 + 240 = 910 torr = 1.20 atm nair = PV/RT = (670 x 1/760)(3.50)/(0.0821)(360) = 0.104mol air => mole fraction CH2Cl2 = nCH2Cl2 / (nair+ nCH2Cl2) = (0.0374)/(0.104 + 0.0374) = 0.265 =26.5% Question 2: nO2 = PV/RT = (11.0)(0.794)/(0.0821)(27 + 273.15) =0.354 mol O2 2C2H2O4(s) + O2(g)---> 4CO2(g) + 2H2O(g) Molar mass C2H2O4 = 2(12.01) +2(1.008) + 4(16.00) = 90.04 g/molC2H2O4 => nC2H2O4 = 8.35/90.04 = 0.0927 molC2H2O4 0.0927 mol C2H2O4 x (1 molO2/2 mol C2H2O4) =0.0464 mol O2 used in reaction. => 0.354 - 0.0464 = 0.308 mol O2 remain in containerafter reaction. 0.0927 mol C2H2O4 x (4 molCO2 / 2 mol C2H2O4) =0.185 mol CO2 produced. 0.0927 mol C2H2O4 x (2 molH2O / 2 mol C2H2O4) =0.0927 mol H2O produced. Ptotal = (nO2 + nCO2 +nH2O)RT/V = (0.308 + 0.185 + 0.0927)(0.0821)(27 +273.15)/(0.794) = 18.2 atm .

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