A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. Which is the limiting reactant and how much excess reactant remains after the reaction
has stopped?

Ammonia is limiting reactant

Amount of oxygen left  = 0.035 mol

Explanation:

Given data:

Masa of ammonia = 2.00 g

Mass of oxygen = 4.00 g

Which is limiting reactant = ?

Excess reactant's amount left = ?

Solution:

Balance chemical equation:

4NH₃ + 3O₂     →     2N₂ + 6H₂O

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 2.00 g/ 17 g/mol

Number of moles = 0.12 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.00 g/ 32 g/mol

Number of moles = 0.125 mol

Now we will compare the moles of ammonia and oxygen with water and nitrogen.

                       NH₃          :            N₂

                         4             :             2

                       0.12           :           2/4×0.12 = 0.06

                       NH₃         :            H₂O

                         4            :             6

                         0.12       :           6/4×0.12 = 0.18

                        O₂            :            N₂

                         3             :             2

                       0.125        :           2/3×0.125 = 0.08

                         O₂           :            H₂O

                         3              :             6

                         0.125       :           6/3×0.125 = 0.25

The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.

Amount of oxygen left:

                         NH₃          :             O₂

                            4            :              3

                            0.12       :          3/4×0.12= 0.09

Amount of oxygen react = 0.09 mol

Amount of oxygen left  = 0.125 - 0.09 = 0.035 mol

1 is the answer to this problem