câu 13 mình giải rồi này bạn xem thử nha, bài 2 :
View attachment 8528
bài 18: PTHH: $2Al+6HCl\rightarrow 2AlCl_{3}+3H_{2}\\a) n_{Al}=\frac{m}{M}=\frac{10,8}{27}=0,4(mol)\\Theo pt: n_{H_{2}}=\frac{3}{2}.n_{Al}=\frac{3}{2}.0,4=0,6(mol)\\\Rightarrow V_{H_{2}}=n.22,4=0,6.22,4=13,44(l)\\b) Theo pt:n_{ctAlCl_{3}}=n_{Al}=0,4(mol)\\\Rightarrow m_{ctAlCl_{3}}=n.M=0,4.133,5=53,4(g)\\Theo pt: n_{ctHCl}=3.n_{Al}=3.0,4=1,2(mol)\\\Rightarrow m_{ctHCl}=n.M=1,2.36,5=43,8(g)\\ \Rightarrow m_{ddHCl}=\frac{m_{ct}}{\frac{C}{100}}.\frac{100}{100}=\frac{43,8}{10,95}.100=400(g)\\ m_{H_{2}}=n.M=0,6.2=1,2(g)\\ \Rightarrow m_{ddAlCl_{3}}=m_{Al}+m_{ddHCl}-m_{H_{2}}=10,8+400-1,2=409,6(g)$
Do đó C%ddAlCl3=mct/mdd.100%=53,4/409,6.100%$\approx $13%