Question

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

Answer 1

Answer:

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 %

Answer 2

Answer:

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 %