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:
[tex]FeActualYield(mol)=\frac{mass(Fe)}{molarmass(Fe)}[/tex]
From the statement, we know that there are 11.6 g of Fe.
We also know the molar mass of the Fe which is:
[tex]molarmass(Fe)=55.845\frac{g}{mol}[/tex]
Now, calculating we have:
[tex]FeActualYield(mol)=\frac{mass(FeProduced)}{molarmass(Fe)}[/tex]
[tex]FeActualYield(mol)=\frac{11.6g}{55.845\frac{g}{mol}}=0.207mol[/tex]
Hence, we have that the actual yield of iron (Fe) in moles is 0.207 mol.
Have a nice day!
Answer:
The actual yield of iron in moles is 34.5%
Explanation:
The balanced chemical equation is
[tex]1Fe_2 O_3+3 C > 2 Fe + 3 CO[/tex]
The conversions are moles [tex]Fe_2 O_3[/tex] to moles Fe and moles Fe to mass Fe
Mole ratio of [tex]Fe_2 O_3:Fe[/tex] is 1 : 2
So using mole ratio we see
[tex]0.301mol Fe_2 O_3 \times \frac {(2mol Fe)}{(1mol Fe_2 O_3 )}[/tex]
=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
[tex]=0.601mol Fe \times 55.85 g/mol[/tex]
=33.566 g Fe produced.
(Theoretical yield)
11.6 g given is the actual yield.
% yield=(actual yield )/(Theoretical yield )×100%
[tex]= \frac {(11.6g)}{(33.6g)} \times 100[/tex]
= 34.5% is the Answer
