Đáp án:

a.  \(m_{Fe}=16,8\ \text{gam}\\ m_{Cu}=3,2\ \text{gam}\)

`\%m_{Fe}=84%, \%m_{Cu}=16%`

b. `m_{CuO}=4\ \text{gam}`

Giải thích các bước giải:

 a. \(Fe+2HCl\to FeCl_2+H_2\\ n_{H_2}=\frac{6,72}{22,4}=0,3 \ \text{mol}\to n_{Fe}=n_{H_2}=0,3\ \text{mol}\to m_{Fe}=0,3.56=16,8\ \text{gam}\to m_{Cu}=20-16,8=3,2\ \text{gam}\)

`-> %m_{Fe}={16,8}/{20}.100\%=84\%-> %m_{Cu}=100%-84%=16%`

b. \(CuO + CO\xrightarrow{t^{\circ}} Cu +CO_2\\ n_{CuO}=n_{Cu}=\frac{3,2}{64}=0,05\ \text{mol}\to m_{CuO}=80.0,05=4\ \text{gam}\)

$n_{H_2}=6,72/22,4=0,3mol$

$PTHH :$

$Fe+2HCl\to FeCl_2+H_2$

$\text{a/Theo pt :}$

$n_{Fe}=n_{H_2}=0,3mol$

$⇒m_{Fe}=0,3.56=16,8g$

$m_{Cu}=20-16,8=3,2g⇒n_{Cu}=3,2/64=0,05mol$

$⇒\%m_{Fe}=\dfrac{16,8.100\%}{20}=84\%$

$⇒\%m_{Cu}=100\%-84\%=16\%$

$b/CuO+CO\overset{t^o}\to Cu+CO_2↑$

$\text{Theo pt :}$

$n_{CuO}=n_{Cu}=0,05mol⇒m_{CuO}=0,06.80=4g$