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.

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.

the answer is definitely c.