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