$Đáp$ $án$ $+$ $Giải$ $thích$ $các$ $bước$ $giải$ $:$

$Câu$ $4$ $:$

$a,$ $PTHH$ $:$

$2Al$ $+$ $6HCl$ $→$ $2AlCl_3$ $+$ $3H_2 ↑$

$b,$

$n_{Al}$ $=$ $\frac{2,7}{27}$ $=$ $0,1$ $(mol)$

$Theo$ $pt$ $:$ $n_{HCl}$ $=$ $3×n_{Al}$ $=$ $0,3$ $(mol)$

$⇒$ $m_{HCl}$ $=$ $0,3×36,5$ $=$ $10,95$ $(g)$

$Theo$ $pt$ $:$ $n_{H_2}$ $=$ $\frac{3}{2}×n_{Al}$ $=$ $0,15$ $(mol)$

$⇒$ $V_{H_2}$ $=$ $0,15×22,4$ $=$ $3,36$ $(lít)$

$c,$

$Cách$ $1$ $:$

$Theo$ $pt$ $:$ $n_{AlCl_3}$ $=$ $n_{Al}$ $=$ $0,1$ $(mol)$

$⇒$ $m_{AlCl_3}$ $=$ $0,1×133,5$ $=$ $13,35$ $(g)$

$Cách$ $2$ $:$

$Áp$ $dụng$ $định$ $luật$ $bảo$ $toàn$ $khối$ $lượng$

$⇒$ $m_{Al}$ $+$ $m_{HCl}$ $=$ $m_{AlCl_3}$ $+$ $m_{H_2}$

$⇒$ $2,7$ $+$ $10,95$ $=$ $m_{AlCl_3}$ $+$ $0,15×2$ 

$⇒$ $m_{AlCl_3}$ $=$ $13,35$ $(g)$

$Chúc$ $bn$ $hk$ $tốt$ $!!$ $:))))$

$Xin$ $ctlhn$ $ạ.$

$a,PTHH:2Al+6HCl\to 2AlCl_3+3H_2\\ b,n_{Al}=\dfrac{2,7}{27}=0,1(mol)\\ \Rightarrow n_{HCl}=3n_{Al}=0,3(mol);n_{H_2}=0,15(mol);n_{AlCl_3}=n_{Al}=0,1(mol)\\ \Rightarrow m_{HCl}=0,3.36,5=10,95(g);V_{H_2}=0,15.22,4=3,36(l)\\ c,C_1:m_{AlCl_3}=0,1.133,5=13,35(g)\\ C_2:m_{H_2}=0,15.2=0,3(g)\\ BTKL:m_{Al}+m_{HCl}=m_{AlCl_3}+m_{H_2}\\ \Rightarrow m_{AlCl_3}=2,7+10,95-0,3=13,35(g)$