What is the percent yield for this reaction under the given conditions? i think i have the right answer(16.6%), but can someone check it for me? !

Percentage yield = 16.4%

which almost equal to 16.6%. So the given answer is true.

Explanation:

Given data:

Mass of hydrogen = 1.52 g

Mass of nitrogen = 9.89 g

Mass of ammonia produced (Actual yield)  = 1.42 g

Percentage yield = ?

Solution:

Chemical equation:

3H₂ + N₂  → 2NH₃

Number of moles of hydrogen:

Number of moles  = mass/ molar mass

Number of moles  = 1.52 g/ 2 g/mol

Number of moles = 0.76 mol

Number of moles of nitrogen:

Number of moles = mass/ molar mass

Number of moles = 9.89 g/ 28 g/mol

Number of moles =  0.35 mol

Now we will compare the moles of NH₃ with nitrogen and hydrogen.

            H₂           :        NH₃

             3             :           2

            0.76         :         2/3 ×0.76 =  0.51 mol

             N₂           :         NH₃

             1              :           2

           0.35          :          2×0.35 = 0.7 mol

The number of moles of NH₃ produced from hydrogen are less that's why Hydrogen will be limiting reactant.

Theoretical yield:

Mass of NH₃ = moles × molar mass

Mass of NH₃ = 0.51 mol × 17 g/mol

Mass of NH₃ = 8.67 g

Theoretical yield of NH₃ = 8.67 g.

Percentage yield:

Percentage yield = actual yield / theoretical yield × 100

Percentage yield =  1.42 / 8.67 × 100

Percentage yield = 16.4%

which almost equal to 16.6%. So the given answer is true.

fe 55.85 g/mol and h2o 18.02 g/mol.