Question

Answer 1

Gọi \(\left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\)

\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\\ n_{Fe_3O_4}=\dfrac{46,4}{232}=0,2\left(mol\right)\)

PTHH:

Zn + H₂SO₄ ---> ZnSO₄ + H₂

a -------------------------------> a

Fe + H₂SO₄ ---> FeSO₄ + H₂

b --------------------------------> b

Hệ pt \(\left\{{}\begin{matrix}65a+56b=37,2\\a+b=0,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,4\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{Zn}=0,4.65=26\left(g\right)\\m_{Fe}=0,2.56=11,2\left(g\right)\end{matrix}\right.\)

PTHH: Fe₃O₄ + 4H₂ --t^o--> 3Fe + 4H₂O

LTL: \(\dfrac{0,2}{1}>\dfrac{0,6}{4}\rightarrow\) Fe₃O₄ dư

Theo pt: \(\left\{{}\begin{matrix}n_{Fe_3O_4\left(pư\right)}=\dfrac{1}{4}n_{H_2}=\dfrac{1}{4}.0,6=0,15\left(mol\right)\\n_{H_2O}=n_{H_2}=0,6\left(mol\right)\\n_{Fe}=\dfrac{3}{4}n_{Fe}=\dfrac{3}{4}.0,6=0,45\left(mol\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{Fe_3O_4\left(dư\right)}=\left(0,2-0,5\right).232=1,16\left(g\right)\\m_{H_2O}=0,6.18=10,8\left(g\right)\\m_{Fe}=0,45.56=25,2\left(g\right)\end{matrix}\right.\)

Answer 2

\(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)
gọi nZn:a  , nFe:b 
=> 65a+56b = 37,2 
pthh: \(Zn+H_2SO_4->ZnSO_4+H_2\)
         a                                           a 
        \(Fe+H_2SO_4->FeSO_4+H_2\)
       b                                         b 
=> a+b = 0,6 
\(\left\{{}\begin{matrix}65a+56b=37,2\\a+b=0,6\end{matrix}\right.=>a=0,4\left(mol\right),b=0,2\left(mol\right)\) 
=> \(m_{Zn}=0,4.65=26\left(g\right)\) 
    \(m_{Fe}=0,2.56=11,2\left(g\right)\)
\(n_{Fe_2O_3}=\dfrac{46,4}{232}=0,2\left(mol\right)\\ pthh:Fe_2O_3+3H_2\underrightarrow{t^O}2Fe+3H_2O\)
LTL: \(\dfrac{0,2}{1}=\dfrac{0,6}{3}\)
=> ko chất nào dư 
=> \(m_{Fe}=0,4.56=22,4\left(g\right)\)