Đáp án:
\(\begin{array}{l} a,\ V=V_{H_2}=6,72\ lít.\\ b,\ C\%_{HCl}=21,9\%\\ c,\ m_{AlCl_3}=26,7\ g.\\ d,\ C\%_{AlCl_3}=25,48\%\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l} a,\ PTHH:2Al+6HCl\to 2AlCl_3+3H_2↑\\ n_{Al}=\dfrac{5,4}{27}=0,2\ mol.\\ Theo\ pt:\ n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\ mol.\\ ⇒V=V_{H_2}=0,3\times 22,4=6,72\ lít.\\ b,\ Theo\ pt:\ n_{HCl}=3n_{Al}=0,6\ mol.\\ ⇒C\%_{HCl}=\dfrac{0,6\times 36,5}{100}\times 100\%=21,9\%\\ c,\ Theo\ pt:\ n_{AlCl_3}=n_{Al}=0,2\ mol.\\ ⇒m_{AlCl_3}=0,2\times 133,5=26,7\ g.\\ d,\ m_{\text{dd spư}}=m_{Al}+m_{\text{dd HCl}}-m_{H_2}\\ ⇒m_{\text{dd spư}}=5,4+100-0,3\times 2=104,8\ g.\\ ⇒C\%_{AlCl_3}=\dfrac{26,7}{104,8}\times 100\%=25,48\%\end{array}\)
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