Giải thích các bước giải:

$a)$

$n_{Al}=\dfrac{5,4}{27}=0,2(mol)$

$n_{HCl}=\dfrac{365×10\%}{36,5}=1(mol)$

$2Al+6HCl\to 2AlCl_3+3H_2$

$\dfrac{n_{Al}}{2}<\dfrac{n_{HCl}}{6}\to HCl$ dư.

$n_{HCl\ \rm (pứ)}=\dfrac{6}{2}n_{Al}=0,6(mol)$

$n_{HCl\ \rm (dư)}=1-0,6=0,4(mol)$

$m_{HCl\ \rm (dư)}=0,4×36,5=14,6(g)$

$b)$

$n_{AlCl_3}=n_{Al}=0,2(mol)$

$n_{H_2}=\dfrac{3}{2}n_{Al}=0,3(mol)$

$m_{AlCl_3}=0,2×133,5=26,7(g)$

$V_{H_2}=0,3×22,4=6,72(l)$

$c)$

$n_{Fe_2O_3}=\dfrac{24}{160}=0,15(mol)$

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

$\dfrac{n_{Fe_2O_3}}{1}>\dfrac{n_{H_2}}{2}\to Fe_2O_3$ dư.

$n_{Fe}=\dfrac{2}{3}n_{H_2}=0,2(mol)$

$n_{Fe_2O_3\ \rm (dư)}=0,15-0,1=0,05(mol)$

$a=m_{Fe}+m_{Fe_2O_3}=0,2×56+0,05×160=19,2(g)$

$d)$

$2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2$

$n_{KMnO_4}=2n_{O_2}=0,6(mol)$

$m_{KMnO_4}=158×0,6=94,8(g)$

Đáp án:

$a\big)HCl\,\text{dư }14,6(g)$

$b\big)m_{AlCl_3}=26,7(g);V_{H_2(đktc)}=6,72(l)$

$c\big)a=19,2(g)$

$d\big)m_{KMnO_4}=47,4(g)$

Giải thích các bước giải:

$a\big)$

$n_{Al}=\frac{5,4}{27}=0,2(mol)$

$n_{HCl}=\frac{365.10\%}{36,5}=1(mol)$

$2Al+6HCl\to 2AlCl_3+3H_2$

Vì $n_{Al}<\frac{n_{HCl}}{3}\to HCl$ dư

Theo PT: $n_{HCl(pứ)}=3n_{Al}=0,6(mol)$

$\to m_{HCl(dư)}=(1-0,6).36,5=14,6(g)$

$b\big)$

Theo PT: $n_{AlCl_3}=n_{Al}=0,2(mol);n_{H_2}=\frac{3}{2}n_{Al}=0,3(mol)$

$\to m_{AlCl_3}=0,2.133,5=26,7(g);V_{H_2(đktc)}=0,3.22,4=6,72(l)$

$c\big)$

$n_{Fe_2O_3}=\frac{24}{160}=0,15(mol)$

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

Vì $n_{Fe_2O_3}>\frac{n_{H_2}}{3}\to Fe_2O_3$ dư

Theo PT: $n_{Fe}=\frac{2}{3}n_{H_2}=0,2(mol);n_{Fe_2O_3(pứ)}=\frac{1}{3}n_{H_2}=0,1(mol)$

$\to n_{Fe_2O_3(dư)}=0,15-0,1=0,05(mol)$

$\to a=m_{CR}=m_{Fe_2O_3(dư)}+n_{Fe}=0,2.56+0,05.160=19,2(g)$

$d\big)$

$2H_2+O_2\xrightarrow{t^o}2H_2O$

Theo PT: $n_{O_2}=\frac{1}{2}n_{H_2}=0,15(mol)$

$2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2$

Theo PT: $n_{KMnO_4}=2n_{O_2}=0,3(mol)$

$\to m_{KMnO_4}=0,3.158=47,4(g)$