Đáp án:
Giải thích các bước giải:
`2Al+6HCl \rightarrow 2AlCl_3+3H_2\uparrow`
a) `n_{Al}=\frac{5,4}{27}=0,2\ mol`
`n_{HCl}=3.n_{Al}=0,6\ mol`
`m_{HCl}=0,6.36,5=21,9\ gam`
`C%_{HCl}=\frac{21,9}{100}.100% = 21,9%`
b) `n_{H_2}=\frac{3}{2}.n_{Al}=0,3\ mol`
`V_{H_2}=0,3.22,4=6,72\ lít`
c) `m_{dd}=m_{Al}+m_{dd \HCl}-m_{H_2}`
`m_{dd}=5,4+100-(0,3.2)=104,8\ gam`
`n_{AlCl_3}=n_{Al}=0,2\ mol`
`C%_{AlCl_3}=\frac{0,2.133,5}{104,8}.100% \approx 25,48%`