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Answer»

Here mass of N2 is 30kg so moles of N2 will be 30*1000/28 = 1071 moles

now mass of H2 is 10kg so no. of moles = 10*1000/2 = 5000moles

the balanced reaction of N2 and H2 forming NH3 is N2 +3H2 → 2NH3

so for 1 mole of N2 3 moles of H2 is required here 1071 moles are there.. so H2 moles required are 1071*3 = 3214

and no. of moles of NH3 formed is 2*(1071) = 2142 moles so, H2 moles will still be there.. after thr complete consumption of N2. remaining moles of H2 = (5000-3214) = 1786 moles

so , the limiting reagent is N2.



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