Combining 0.350 mol Fe 2 O 3 with excess carbon produced 18.9 g Fe. Fe 2 O 3 + 3 C ⟶ 2 Fe + 3 CO

What is the actual yield of iron in moles?

What is the theoretical yield of iron in moles?

What is the percent yield?

1. 0.338 moles of Fe

2. 0.700 moles of Fe

3. 48.3%

Explanation:

This is the reaction:

Fe₂O₃ + 3C → 2Fe + 3CO

We were told that we produce 18.9 g of Fe. Let's convert the mass to moles:

18.9 g. 1mol / 55.85 g = 0.338 moles of Fe

Let's make a rule of three; ratio is 1:2.

1 mol of oxide can produce 2 moles of elemental iron

Then, 0.350 moles must produce (0.350.2) / 1 = 0.700 moles of Fe

Let's determine the percent yield:

(Yield produced / Theoretical Yield). 100 = 48.3 %

The actual yield of iron in moles = 0.338 moles

The theoretical yield of iron in moles = 0.700 moles iron

The percentage yield = 48.3 %

Explanation:

Step 1: Data given

Number of moles Fe2O3 = 0.350 moles

carbon = in excess

Mass of Fe = 18.9 grams

Step 2: The balanced equation

Fe2O3 + C ⟶ 2 Fe + 3 CO

Step 3: Calculate moles Fe

For 1 mol Fe2O3 we need 1 mol C to produce 2 moles Fe and 3 moles CO

For 0.350 moles Fe2O3 we'll have 2*0.350 = 0.700 moles Fe

Step 4: Calculate mass Fe

Mass Fe = moles Fe * molar mass Fe

Mass Fe = 0.700 moles * 55.845 g/mol

Mass Fe = 39.10 grams = the theoretical yield

Step 5: Calculate % yield

% yield = (actual yield / theoretical yield) * 100%

% yield = (18.9 grams / 39.10 grams) * 100%

% yield = 48.3 %

Step 6: Calculate moles Fe actual yield

Moles Fe = 18.9 grams / 55.845 g/mol

Moles Fe = 0.338 moles

% yield in moles = (0.338 moles / 0.700 moles) * 100 %

% yield in moles = 48.2 %