10 MOLES N2 AND 15 MOLES H2 WERE ALLOWED TO REACT OVER A SUITABLE CATALYST .10 MOLES OF NH3 WERE FORMED .THE REMAINING MOLES OF N2 AND H2 RESPECTIVELY ARE:

1) 5 MOLES ,0 MOLES (2) 0 MOLES,5 MOLES

3) 9 MOLES,12 MOLES (4)0 MOLES, 0 MOLES

SIR / MADAM,COULD YOU PLEASE SOLVE THE ENTIRE SUM,SO THAT I GET TO KNOW HOW IT IS SOLVED,AND THEN CHOOSE THE CORRECT OPTION.

_{2}and H

_{2}to form NH

_{3}is given as,

N

_{2}+ 3H2 $\to $ 2NH3

According to the equation, 1 mole of N

_{2}requires 3moles of H

_{2}to form 2 moles of NH

_{3}.

Ratio- N

_{2}: H

_{2}: NH

_{3}= 1 : 3 : 2

Now we have 10 moles of N

_{2}for which we require 3$\times $10 = 30 moles of H

_{2}to react completely. But we have only 15 moles of H

_{2}that means H

_{2}is the limiting reactant.

Therefore, 15 moles of H

_{2} will react only with 5 moles of N

_{2}to form 10 moles of NH

_{3}.

So correct option is 1) 5 moles of N

_{2}will remain and 0 moles of H

_{2}will remain.

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