Question

cho mik xin đáp án vs ạ

 

Answer 1

Gọi: \(\left\{{}\begin{matrix}n_{CuO}=a\left(mol\right)\\n_{MgO}=b\left(mol\right)\\n_{FeO}=c\left(mol\right)\end{matrix}\right.\)

- Khi cho pư với H₂.

PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

\(FeO+H_2\underrightarrow{t^o}Fe+H_2O\)

Ta có: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{CuO}+n_{FeO}=a+c=0,5\left(1\right)\)

\(\left\{{}\begin{matrix}n_{Cu}=n_{CuO}=a\left(mol\right)\\n_{Fe}=n_{FeO}=c\left(mol\right)\end{matrix}\right.\)

⇒ 64a + 56c = 29,6 (2)

- Cho hh pư với dd HCl.

PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

\(FeO+2HCl\rightarrow FeCl_2+H_2\)

Ta có: \(n_{HCl}=\dfrac{43,8}{36,5}=1,2\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{CuO}+2n_{MgO}+2n_{FeO}=2a+2b+2c=1,2\left(3\right)\)

Từ (1), (2) và (3) \(\Rightarrow\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,1\left(mol\right)\\c=0,3\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CuO}=0,2.80=16\left(g\right)\\m_{MgO}=0,1.40=4\left(g\right)\\m_{FeO}=0,3.72=21,6\left(g\right)\end{matrix}\right.\)