Question

Answer 1

a, Gọi \(\left\{{}\begin{matrix}n_{CuO}=a\left(mol\right)\\n_{Fe_2O_3}=b\left(mol\right)\end{matrix}\right.\left(đk:a,b>0\right)\)

PTHH:

$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$

  a------------>a

$Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O$

  b----------->2b

Theo bài ra, ta có hệ: \(\left\{{}\begin{matrix}80a+160b=9,6\\64a+2a.56=7,04\end{matrix}\right.\)

\(\Leftrightarrow a=b=0,04\left(mol\right)\left(TM\right)\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{Fe_2O_3}=\dfrac{0,04.160}{9,6}.100\%=66,67\%\\\%m_{Cu}=100\%-66,67\%=33,33\%\end{matrix}\right.\)

b, PTHH: $Fe + H_2SO_4 \rightarrow FeSO_4 + H_2 \uparrow$

               0,08----->0,08----------------->0,08

\(\rightarrow m_{ddH_2SO_4}=\dfrac{0,08.98}{29,4\%}=\dfrac{80}{3}\left(g\right)\)

\(c,V_{H_2}=0,08.22,4=1,792\left(l\right)\)