Question

What is the percentage of the actual yield of the reaction of 2.5 mol of Fe(NO3),

Answer 1

Answer:

Percent yield = 87.5%

Explanation:

Given data:

Number of moles of Fe(NO₃)₃ = 2.5 mol

Number of moles of Na₂CO₃ = 3.6 mol

Actual yield of NaNO₃ = 6.3 mol

Percent yield of NaNO₃ = ?

Solution:

Chemical equation:

2Fe(NO₃)₃ + 3Na₂CO₃        →     Fe₂(CO₃)₃ + 6NaNO₃

now we will compare the moles of both reactant with NaNO₃.

                   Fe(NO₃)₃          :              NaNO₃

                       2                   :                  6

                     2.5                  :            6/2×2.5 = 7.5

                  Na₂CO₃             :              NaNO₃

                       3                   :                  6

                     3.6                  :            6/3×3.6 = 7.2

Theoretical yield of NaNO₃ is 7.2 mol.

Percent yield:

Percent yield = actual yield / theoretical yield ×100

Percent yield = 6.3 mol / 7.2mol ×100

Percent yield = 87.5%