Đáp án + Giải thích các bước giải:

Câu 3:

$a\big)$

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

$\to A:Fe,Al_2O_3$

$b\big)$

$n_{H_2O}=\frac{4,05}{18}=0,225(mol)$

Theo PT: $n_{Fe_2O_3}=0,075(mol)$

$\to\%m_{Fe_2O_3}=\frac{0,075.160}{16}.100\%=75\%$

$\to \%m_{Al_2O_3}=100-75=25\%$

Câu 4:

$n_{Fe}=\frac{33,6}{56}=0,6(mol)$

$Fe_3O_4+4H_2\xrightarrow{t^o}3Fe+4H_2O$

Theo PT: $n_{H_2}=0,8(mol);n_{Fe_3O_4}=0,2(mol)$

$\to V=0,8.22,4=17,92(l)$

$\to m_{Fe_3O_4}=0,2.232=46,4(g)$

Câu 5:

$a\big)2Zn+O_2\xrightarrow{t^o}2ZnO$

$b\big)$

$n_{O_2}=\frac{13,44.20\%}{22,4}=0,12(mol)$

Theo PT: $n_{Zn}=n_{ZnO}=0,24(mol)$

$\to m=m_{Zn}=0,24.65=15,6(g)$

$\to m_1=m_{ZnO}=0,24.81=19,44(g)$

Câu 6:

$a\big)$

$n_{CH_4}=\frac{12}{16}=0,75(mol)$

$n_{O_2}=\frac{11,2}{22,4}=0,5(mol)$

$CH_4+2O_2\xrightarrow{t^o}CO_2+2H_2O$

Vì $n_{CH_4}>\frac{n_{O_2}}{2}\to CH_4$ dư

$b\big)$

Theo PT: $n_{CO_2}=\frac{1}{2}n_{O_2}=0,25(mol)$

$\to V_{CO_2(đktc)}=0,25.22,4=5,6(l)$