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What is the percentage of the actual yield of the reaction of 2.5 mol of Fe(NO3),
with 3.6 mol of Na2Co3, to form 6.3 mol of NaNO3 according to the reaction :
2Fe(NO3)3 + 3NaC03 - >
Fe2(CO3)3 + 6NaNO3 ?​

Sagot :

Neetooidn

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%

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