Identify limiting and excess reagents when 25g of nitrogen reacts with 25g of hydrogen. How many grams of ammonia gas are formed in this reaction?

The mass of ammonia produced in the reaction is  30.26 g.

The equation of the reaction is;

N2 (g) + 3H2(g) > 2NH3(g)

Number of moles nitrogen = 25g /28 g/mol = 0.89 moles

Number of moles of hydrogen =  25g /2 g/mol = 12.5 moles

1 mole of N2 reacts with 3 moles of H2

0.89 moles of N2 reacts with 0.89 moles ×  3 moles/1 mole = 2.67 moles of H2

We can see that there is more than enough H2 present in the system hence the H2 is the reactant in excess while N2 is the limiting reactant.

If 1 mole of N2 yields 2 moles of NH3

0.89 moles of N2 yields 0.89 moles ×  2 moles /1 mole = 1.78 moles of NH3

Mass of NH3 produced = 1.78 moles of NH3 × 17 g/mol = 30.26 g

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Nitrogen is limiting reactant while hydrogen is in excess.

Explanation:

Given data:

Mass of N₂ = 25 g

Mass of H₂ = 25 g

Mass of ammonia formed = ?

Solution:

Chemical equation:

N₂ + 3H₂    →     2NH₃

Number of moles of Nitrogen:

Number of moles = mass/ molar mass

Number of moles = 25 g/ 28 g/mol

Number of moles = 0.89 mol

Number of moles of hydrogen:

Number of moles = mass/ molar mass

Number of moles = 25 g/ 2 g/mol

Number of moles = 12.5 mol        

Now we will compare the moles of both reactant with ammonia.

                   H₂            ;             NH₃

                    3             :              2

                    12.5        :            2/3×12.5 = 8.3

                   N₂            ;             NH₃

                    1              :              2

                    0.89        :            2×0.89 = 1.78

The number of moles of ammonia produced by nitrogen are less thus nitrogen is limiting reactant while hydrogen is in excess.

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