FeO + H2SO4\(\rightarrow\)FeSO4 + H2 (1)
CuO + H2SO4\(\rightarrow\)CuSO4 + H2 (2)
a;
\(m_{H_2SO_4}=100\dfrac{19,6}{100}=19,6\left(g\right)\)
Đặt \(n_{FeO}\)là a
\(n_{CuO}\)là b
Ta có hệ pt:
\(\left\{{}\begin{matrix}72a+80b=15,2\\98a+98b=19,6\end{matrix}\right.\)
Giải hệ pt ta có:
a=0,1;b=0,1
\(m_{FeO}=72.0,1=7,2\left(g\right)\)
\(m_{CuO}=0,1.80=8\left(g\right)\)
b;
Theo PTHH 1 và 2 ta có:
\(n_{FeO}=n_{FeSO_4}=0,1\left(mol\right)\)
\(n_{CuO}=n_{CúSO_4}=0,1\left(mol\right)\)
\(m_{FeSO_4}=154.0,1=15,4\left(g\right)\)
\(m_{CuSO_4}=162.0,1=16,2\left(g\right)\)
\(C\%\)dd FeSO4 là:\(\dfrac{15,4}{100+15,2-0,1.2}.100\%\approx13,4\%\)
C% dd CuSO4 là:\(\dfrac{16,2}{100+15,2-0,1.2}\approx14\%\)