Bài 2:
$a)n_{Mg}=\dfrac{3,6}{24}=0,15(mol)$
$Mg+2HCl\to MgCl_2+H_2$
Lập tỉ lệ: $\dfrac{n_{Mg}}{1}>\dfrac{n_{HCl}}{2}\Rightarrow Mg$ dư
$\Rightarrow n_{H_2}=0,5.n_{HCl}=0,075(mol)$
$\Rightarrow V_{H_2}=0,075.22,4=1,68(l)$
$b)n_{Mg(dư)}=0,15-0,5.0,15=0,075(mol)$
Theo PT: $n_{MgCl_2}=n_{H_2}=0,075(mol)$
$\Rightarrow m_{MgCl_2}=0,075.95=7,125(g)$
$m_{Mg(dư)}=0,075.24=1,8(g);m_{H_2}=0,075.2=0,15(g)$
Bài 3:
$a)n_{Al}=\dfrac{5,4}{27}=0,2(mol)$
$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$
Lập tỉ lệ: $\dfrac{n_{Al}}{2}<\dfrac{n_{H_2SO_4}}{3}\Rightarrow H_2SO_4$ dư
$\Rightarrow n_{H_2}=1,5.n_{Al}=0,3(mol)$
$\Rightarrow V_{H_2}=0,3.22,4=6,72(l)$
$b)n_{H_2SO_4(dư)}=0,6-0,3=0,3(mol)$
Theo PT: $n_{Al_2(SO_4)_3}=0,5.n_{Al}=0,1(mol)$
$\Rightarrow m_{H_2SO_4(dư)}=0,3.98=29,4(g)$
$m_{H_2}=0,3.2=0,6(g);m_{Al_2(SO_4)_3}=0,1.342=34,2(g)$
