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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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                  

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.

the correct answer would be d. hope this !