Câu 4:

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

$a)PTHH:2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\uparrow(1)$

$ZnO+H_2\xrightarrow{t^o}Zn+H_2O(2)$

$b)$ Theo PT(1): $n_{H_2SO_4}=1,5n_{Al}=0,3(mol)$

$\Rightarrow C\%_{H_2SO_4}=\dfrac{0,3.98}{150}.100\%=19,6\%$

$c)$ Theo PT(1): $n_{H_2}=n_{H_2SO_4}=0,3(mol)$

Theo PT(2): $n_{Zn}=n_{H_2}=0,3(mol)$

$\Rightarrow m_{Zn}=0,3.65=19,5(g)$

$\to m=19,5$

Câu 5:

$2K+2H_2O\to 2KOH+H_2\uparrow$

$4Al+3O_2\xrightarrow{t^o}2Al_2O_3$

$2Fe_xOy+\dfrac{3x-2y}{2}O_2\to xFe_2O_3$

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

Câu 7:

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

$a)PTHH:2Al+6HCl\to 2AlCl_3+3H_2$

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

Đổi $400ml=0,4l$

$\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,4}=1,5M$

$b)$ Theo PT: $n_{AlCl_3}=n_{Al}=0,2(mol)$

$\Rightarrow m_{AlCl_3}=0,2.133,5=26,7(g)$

$c)$ Theo PT: $n_{H_2}=1,5n_{Al}=0,3(mol)$

$\Rightarrow V_{H_2}=0,3.22,4=6,72(lít)$