Question

Combining 0.380 mol Fe2O3 with excess carbon produced 13.7 g Fe

Answer 1

Answer:

Percent yield = 32.9%

Explanation:

Given data:

Number of moles of Fe₂O₃ = 0.380 mol

Number of moles of Fe produced = 13.7 g

Theoretical yield = ?

Actual yield of iron in mol = ?

Percent yield = ?

Solution:

Chemical equation:

Fe₂O₃ + 3C   →   2Fe  + 3CO

Actual yield of iron in moles:

Number of  moles = mass/ molar mass

Number of moles = 13.7 g/ 55.8 g/mol

Number of moles = 0.25 mol

Theoretical yield:

                    Fe₂O₃          :           Fe

                      1                :              2

                    0.380         :          2×0.380 = 0.76

Theoretical yield = 0.76 mol

Percent yield:

Percent yield = actual yield / theoretical yield × 100

Percent yield = 0.25 mol /0.76 mol × 100

Percent yield = 32.9%