How do I go about solving this? A mixture of N2(g) and H2(g) reacts in a closed

Question

How do I go about solving this? A mixture of N2(g) and H2(g) reacts in a closed container toform ammonia NH3(g). The reaction ceases before either reactant hasbeen totally consumed. At this stage 3.0 moles of N2, 3.0 moles ofH2 and 3.0 moles of NH3 are present. How many moles of N2, and H2were present originally? How do I go about solving this? A mixture of N2(g) and H2(g) reacts in a closed container toform ammonia NH3(g). The reaction ceases before either reactant hasbeen totally consumed. At this stage 3.0 moles of N2, 3.0 moles ofH2 and 3.0 moles of NH3 are present. How many moles of N2, and H2were present originally?

Explanation / Answer

A mixture of N2(g) and H2(g) reacts in a closed container toform ammonia NH3(g). The reaction ceases before either reactant hasbeen totally consumed. At this stage 3.0 moles of N2, 3.0 moles ofH2 and 3.0 moles of NH3 are present. How many moles of N2, and H2were present originally                                    N2(g) + 3H2(g) -------> 2NH3(g) 1 mole of N2 reacts with 3 moles of H2 produces 2 moles of NH3 To produce 3 moles of NH3 we require ( 1 * 3 )/ 2   moles of N 2 and    ( 3 *3 ) / 2 moles of H 2 i.e., 3 moles of NH3 we require 1.5   moles ofN 2 and    4.5 moles of H 2 But in the problem we have  3.0 moles ofNH3 , 3.0 moles of N2 and 3.0 moles of H2 No. of moles of N2 initially present = moles remained +reacted moles                                                      = 3 + 1.5                                                       =4.5 moles No. of moles of H2 initially present = molesremained + reacted moles                                                       = 3.0 + 4. 5                                                       = 7.5 moles

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