Question

Combining 0.301 mol Fe2O3

with excess carbon produced 11.6 g Fe.


Fe2O3+3C⟶2Fe+3CO


What is the actual yield of iron in moles?

Answer 1

The actual yield of iron in moles is 34.5%

Step-by-step explanation:

The balanced chemical equation is

`1Fe_2 O_3+3 C > 2 Fe + 3 CO`

The conversions are moles
`Fe_2 O_3` to moles Fe and moles Fe to mass Fe

Mole ratio of
`Fe_2 O_3:Fe` is 1 : 2

So using mole ratio we see

`0.301mol Fe_2 O_3  * \frac {(2mol Fe)}{(1mol Fe_2 O_3 )}`

=0.601 mol Fe is produced.

By multiplying with molar mass of Fe we can convert moles Fe to mass Fe.

That is

Mass Fe = moles Fe × molar mass Fe

`=0.601mol Fe * 55.85 g/mol`

=33.566 g Fe produced.

(Theoretical yield)

11.6 g given is the actual yield.

% yield=(actual yield )/(Theoretical yield )×100%

`= \frac {(11.6g)}{(33.6g)} * 100`

= 34.5% is the Answer

Answer 2

The actual yield of iron (Fe) in moles is 0.207 mol.

Why?

Since we know the chemical equation, and how much Fe is produced, we can calculate the actual yield of iron in moles using the following equation:

`FeActualYield(mol)=(mass(Fe))/(molarmass(Fe))`

From the statement, we know that there are 11.6 g of Fe.

We also know the molar mass of the Fe which is:

`molarmass(Fe)=55.845(g)/(mol)`

Now, calculating we have:

`FeActualYield(mol)=(mass(FeProduced))/(molarmass(Fe))`

`FeActualYield(mol)=(11.6g)/(55.845(g)/(mol))=0.207mol`

Hence, we have that the actual yield of iron (Fe) in moles is 0.207 mol.

Have a nice day!