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Đáp án + Giải thích các bước giải:

Bài 5:

$Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O$

$3Fe+2O_2\xrightarrow{t^o}Fe_3O_4$

$n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1(mol)$

Theo PT: $n_{Fe}=3n_{Fe_3O_4}=0,3(mol)$

$\to m_{Fe}=0,3.56=16,8(g)$

Theo PT: $n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=0,15(mol)$

$\to m_{Fe_2O_3}=0,15.160=24(g)$

$\to a=24;b=16,8$

Bài 6:

$CuO+H_2\xrightarrow{t^o}Cu+H_2O$

$Fe_2O_3+3H_2\xrightarrow{t^o}2Fe+3H_2O$

$a)V_{H_2(đktc)}=5,04(l)$

$b)n_{H_2}=\dfrac{5,04}{22,4}=0,225(mol)$

$\to m_{H_2}=0,225.2=0,45(g)$

Theo PT: $n_{H_2O}=n_{H_2}=0,225(mol)$

$\to m_{H_2O}=0,225.18=4,05(g)$

Bảo toàn khối lượng ta có:

$m_{\rm oxit}+m_{H_2}=m_{\text{kim loại}}+m_{H_2O}$

$\to m_{\text{kim loại}}=14+0,45-4,05=10,4(g)$

$c)Zn+2HCl\to ZnCl_2+H_2\uparrow$

Theo PT: $n_{Zn}=n_{H_2}=0,225(mol);n_{HCl}=2n_{H_2}=0,45(mol)$

$\to \begin{cases} m_{Zn}=0,225.65=14,625(g)\\ m_{HCl}=0,45.36,5=16,425(g) \end{cases}$