Question

In the reaction given below, 23.5 grams of iron(III) oxide (Fe2O3) are reacted with excess carbon. What is the theoretical yield of iron metal?

2 Fe2O3 + 3 C --> 4 Fe + 3 CO2

If 14.5 grams of iron are recovered when the reaction is complete, what is the percent yield?


**Both answer should be written as XX.X

Answer 1

I tried the answer above and it’s correct

Step-by-step explanation:

Answer 2

Theoretical yield = 16.76 g

Percent yield = 87%

Step-by-step explanation:

Given data:

Mass of iron oxide = 23.5 g

Theoretical yield of iron = ?

Actual yield of iron = 14.5 g

Percent yield of iron = ?

Solution:

Chemical equation:

2Fe₂O₃ + 3C → 4Fe + 3CO₂

Number of moles of iron oxide:

Number of moles = mass/molar mass

Number of moles = 23.5 g/ 159.69 g/mol

Number of moles = 0.15 mol

Now we will compare the mole of iron oxide with iron metal.

Fe₂O₃ : Fe

2 : 4

0.15 : 4/2×0.15 = 0.3

Mass of iron:

Mass = number of moles × molar mass

Mass = 0.3 mol × 55.85 g/mol

Mass = 16.76 g

Percent yield:

percent yield = (actual yield / theoretical yield)× 100

Percent yield = (14.5 g / 16.76 g) × 100

Percent yield = 0.87 × 100

Percent yield = 87%