Gaseous ammonia, NH3, canbe prepared by the reaction of gaseous hydrogen and gas
ID: 685449 • Letter: G
Question
Gaseous ammonia, NH3, canbe prepared by the reaction of gaseous hydrogen and gaseousnitrogen. 3H2 (g) +N2 (g) -->2NH3 (g) A 150 L flask containing 12.0 g of H2(MH2 = 2g/mol) is charged with 28.0 g of N2(MN2 =28 g/mol) and the reaction proceeds until one of the reactants iscompletely consumed (MNH3 = 17 g/mol). The temperature in the flask is 125C. a) What is the partial pressure in atm, of N2 in the flask after the reaction iscompleted? b)What is the partial pressure in atm, of NH3 in the flask after the reaction iscompleted? c)What is the partial pressure in atm, of H2 in the flask after the reaction iscompleted? d)What is the total pressure in atm, in the flask after thereaction is completed? e)Calculate the final composition of the gas in the flask inmole fractions using the Dalton Law I got the partial pressure of H2 by: 12g/2g mol = 6 mol H2 T = 125C + 273.15 = 398.15 K partial pressure of H2 = 6 mol X 0.0821 L atm X 398.15/150 L =1.308 atm for N2: 28g/28g/mol = 1 mol partial pressure of N2 = 1 mol X 0.0821 L atm X398.15/150 L = 0.218 atm then 12g + 28g = 40g 40g/17g mol = 2.35 mol of NH3 partial pressure of NH3 = 2.35 mol X 0.0821 L atm X 398.15/150 L = 0.512 atm total pressure = 1.308 atm + 0.218 atm + 0.512 atm = 2.038atm total mol = 9.35 mol for mole fraction: H2: 6 mol/ 9.35 mol = 0.64 N2: 1 mol/9.35 mol = 0.11 HN3 = 2.35 mol/9.35 mol = 0.25 0.64 + 0.11 + 0.25 = 1 mole fraction Professor said my answers were wrong but I am not surewhy. Please help Gaseous ammonia, NH3, canbe prepared by the reaction of gaseous hydrogen and gaseousnitrogen. 3H2 (g) +N2 (g) -->2NH3 (g) A 150 L flask containing 12.0 g of H2(MH2 = 2g/mol) is charged with 28.0 g of N2(MN2 =28 g/mol) and the reaction proceeds until one of the reactants iscompletely consumed (MNH3 = 17 g/mol). The temperature in the flask is 125C. a) What is the partial pressure in atm, of N2 in the flask after the reaction iscompleted? b)What is the partial pressure in atm, of NH3 in the flask after the reaction iscompleted? c)What is the partial pressure in atm, of H2 in the flask after the reaction iscompleted? d)What is the total pressure in atm, in the flask after thereaction is completed? e)Calculate the final composition of the gas in the flask inmole fractions using the Dalton Law I got the partial pressure of H2 by: 12g/2g mol = 6 mol H2 T = 125C + 273.15 = 398.15 K partial pressure of H2 = 6 mol X 0.0821 L atm X 398.15/150 L =1.308 atm for N2: 28g/28g/mol = 1 mol partial pressure of N2 = 1 mol X 0.0821 L atm X398.15/150 L = 0.218 atm then 12g + 28g = 40g 40g/17g mol = 2.35 mol of NH3 partial pressure of NH3 = 2.35 mol X 0.0821 L atm X 398.15/150 L = 0.512 atm total pressure = 1.308 atm + 0.218 atm + 0.512 atm = 2.038atm total mol = 9.35 mol for mole fraction: H2: 6 mol/ 9.35 mol = 0.64 N2: 1 mol/9.35 mol = 0.11 HN3 = 2.35 mol/9.35 mol = 0.25 0.64 + 0.11 + 0.25 = 1 mole fraction Professor said my answers were wrong but I am not surewhy. Please help d)What is the total pressure in atm, in the flask after thereaction is completed? e)Calculate the final composition of the gas in the flask inmole fractions using the Dalton Law I got the partial pressure of H2 by: 12g/2g mol = 6 mol H2 T = 125C + 273.15 = 398.15 K partial pressure of H2 = 6 mol X 0.0821 L atm X 398.15/150 L =1.308 atm for N2: 28g/28g/mol = 1 mol partial pressure of N2 = 1 mol X 0.0821 L atm X398.15/150 L = 0.218 atm then 12g + 28g = 40g 40g/17g mol = 2.35 mol of NH3 partial pressure of NH3 = 2.35 mol X 0.0821 L atm X 398.15/150 L = 0.512 atm total pressure = 1.308 atm + 0.218 atm + 0.512 atm = 2.038atm total mol = 9.35 mol for mole fraction: H2: 6 mol/ 9.35 mol = 0.64 N2: 1 mol/9.35 mol = 0.11 HN3 = 2.35 mol/9.35 mol = 0.25 0.64 + 0.11 + 0.25 = 1 mole fraction Professor said my answers were wrong but I am not surewhy. Please helpExplanation / Answer
The balanced equation of the formation ofammonia is N2 (g) + 3H2 --------> 2 NH3 (g) From the given information wehave, The number of moles of hydrogeninitially = 12g/2g /mol = 6 mol H2 The number of moles of nitrogen added =28.0 g/ 28.0 g/mol = 1 mol But according to givenreaction, 1 mole of H2 reacted with the3 moles of H2. Hence - From the above values, we conclude thatN2 is a limiting reactant. Therefore the number of moles ofNH3 present = 1 mol N2 *(2 molNH3/ 1 mol N2) = 2 mol NH3 Number of moles of H2 consumed= 3 moles Therefore number of moles ofH2 remained = 3 moles Number of moles of N2 remained= 0 mol (limiting reactant) Therefore total moles of container afterreaction was complete = 2 mol NH3 + 3 molesH2 =5 moles Thus we have n = 5 mol R = 0.0821 L.atm/mol.K V = 150 L =5 moles Thus we have n = 5 mol R = 0.0821 L.atm/mol.K V = 150 L T = 125C + 273.15 = 398.15 K Therefore - Total pressure after completion of the reaction =5*0.0821*398.15/150 = 1.089 atm Mole fraction of the H2 afterreaction is complete = 3/5 = 0.6 Mole fraction of the NH3 = 2/5 = 0.4 Therefore partial pressure ofNH3 = 0.4* total pressure = 0.4 *1.089 atm = 0.435 atm Partial pressure of the H2 =0.6*1.089 atm = 0.6534 atmRelated Questions
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