Question

Mong Được các ad giúp đỡ

 

Answer 1

a, 

PTHH: CuO + H₂SO₄ → CuSO₄ + H₂O

Mol:       x             x                 x

PTHH: Fe₂O₃ + 3H₂SO₄ → Fe₂(SO₄)₃ + 3H₂

Mol:        y              3y                    y

Ta có: \(\left\{{}\begin{matrix}80x+160y=56\\160x+400y=136\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,3\end{matrix}\right.\)

\(m_{CuO}=0,1.80=8\left(g\right);m_{Fe_2O_3}=56-8=48\left(g\right)\)

b, \(\%m_{CuO}=\dfrac{8.100\%}{56}=14,29\%\)

\(\%m_{Fe_2O_3}=\dfrac{48.100\%}{56}=85,71\%\)

c, \(n_{H_2SO_4}=n_{CuO}+3n_{Fe_2O_3}=0,1+3.0,3=1\left(mol\right)\)

\(\Rightarrow C_{M_{ddH_2SO_4}}=\dfrac{1}{0,2}=5M\)