Đáp án:

4. a. $24\text{(g)}$    b. $16\text{(g)}$

5. a. $28\text{(g)}$    b. $7,467\text{(l)}$

Giải thích các bước giải:

4. \(a)\; n_{H_2}=\dfrac{6,72}{22,4}=0,3\text{(mol)}\\ PTHH:\; CuO + H_2\xrightarrow{t^{\circ}} Cu+ H_2O\\ n_{CuO}=n_{H_2}=0,3\text{(mol)}\to m_{CuO}=0,3\times 80=24\text{(g)} \\ b)\; n'_{H_2}=\dfrac{4,48}{22,4}=0,2\text{(mol)}\to n'_{CuO}=n_{H_2}=0,2\text{(mol)}\to m'_{CuO}=0,2\times 80 = 16\text{(g)}\)

5. $PTHH: \; FeO+H_2\xrightarrow{t^{\circ}}Fe+H_2O\\ a)\;n_{Fe}=\dfrac{36}{72}=0,5\text{(mol)}\to n_{Fe}=n_{FeO}=0,5\text{(mol)}\to m_{Fe}=56\times 0,5=28\text{(g)}\\ b)\; PTHH:\;3Fe+ 2O_2\xrightarrow{t^{\circ}}Fe_3O_4\\n_{O_2}=\dfrac 23\times n_{Fe}=\dfrac 23\times 0,5=\dfrac 13\text{(mol)}\to V_{O_2}=22,4\times \dfrac 13=7,467\text{(l)}$