\(\begin{array}{l}
a)\\
1\,tan=1000\,kg\\
m_{Fe_3O_4}=1000.65,71\%=657,1(kg)\\
\to n_{Fe_3O_4}=\frac{657,1}{232}\approx 2,832(g)\\
\to n_{Fe}=3n_{Fe_3O_4}=8,496(g)\\
\to m_{Fe}=8,496.56=475,776(g)\\
b)\\
1\,tan=1000\,kg\\
n_{Fe}=\frac{1000}{56}=\frac{125}{7}(mol)\\
\to n_{Fe_3O_4}=\frac{1}{3}n_{Fe}=\frac{125}{21}(mol)\\
\to m_{quang}=\dfrac{\frac{125}{21}.232}{65,71\%}\approx 2102(kg)=2,102(tan)
\end{array}\)

Đáp án:

`a.\ m_(Fe)≈475,806(kg)`

`b.\ m_(text(quặng))≈2,13(text(tấn))`

Giải thích các bước giải:

`a.`

`%m_(Fe//Fe_3O_4)=(56xx3)/232xx100≈72,41%`

`m_(Fe_3O_4)=1xx65,71%=0,6571(text(tấn))`

`=>m_(Fe)=0,6571xx71,41%≈0,475806(text(tấn))=475,806(kg)`

`b.`

`m_(Fe_3O_4)=1:71,41%≈1,4(text(tấn))`

`=>m_(text(quặng))=1,4:65,71%≈2,13(text(tấn))`