For the reaction n2 + 3h2 = 2nh3, if you have 2.00 moles of n2 and 4.00 moles of h2 : a. identify the limiting reactant. b. how many moles of nh3 can be formed? c. how many moles of the excess reactant will remain?
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The answer to your question is below

Explanation:

For the reaction N2 + 3H2 = 2NH3, if you have 2.00 moles of N2 and 4.00 moles of H2 : a. Identify the limiting reactant.

                                   N2     +     3H2      =      2NH3

                                 1 mol         3 moles        2 moles

Given                       2 moles     4 moles

       Rule of three for N2

                    x =  (2 x 3) / 1 = 6 mol  of H2, it means that is needed 6 moles of H2 and there are only 4, so the limiting reactant is H2                                

b. How many moles of NH3 can be formed?

           rule of three

                                3 moles of H2  2 moles of NH3              

                                4 moles of H2     x

                     x = (4x2)/3 = 2.7 moles of NH3 are formed

c. How many moles of the excess reactant will remain?

      x = (4 x 1) / 3     =  1.33

Excess reactant = 2 - 1.33 = 0.67 moles                  

the answer is 75.0%.