InterviewSolution
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InterviewSolution
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50.0 kg of N, (g) and 10.0 kg of H, (g) aremixed to produce NH, (g). Calculate theNH, (g) formed. Identify the limitingreagent in the production of NH, in thissituation. |
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Answer» Let us write the balanced equationN2 + 3H2 → 2NH3Now calculate the number of molesNumber of moles of N2 = 50 kg of N2 = 50 X 10 3 g/1 kg x 28g = 17.86 x 10 2 mole Number of moles of H2 = 10 kg of N2 = 10 X 103 g/ 1 kg x 2 = 4.96X 103 mol According to the above equation 1 mole of N2 reacts with 3 moles H2.That is 17.86 x 10 2 mole of N2 reacts with ------moles of H2= 3/1 X 17.86 x 10 2 = 5.36 x 103 moles.Here we have 4.96X 103 mol of hydrogen. Hence Hydrogen is the limiting reagent.Let us calculate the amount ammonia formed by reacting 4.96X103 moles Hydrogen 3 moles of hydrogen -------2 moles of NH34.96 x103 moles Hydrogen -----?= 4.96 x103 X ⅔= 3.30 x 103 moles of NH3 |
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