Đáp án:

Câu 1:

\(\begin{array}{l} a,\ V_{H_2}=6,72\ lít.\\ b,\ m_{AlCl_3}=26,7\ g.\\ c,\ C_{M_{HCl}}=3\ M.\\ d,\ C_{M_{AlCl_3}}=1\ M.\end{array}\)

Câu 2:

\(\begin{array}{l} a,\ V_{H_2}=4,48\ lít.\\ b,\ C\%_{\text{dd spư}}=15,15\%\end{array}\)

Câu 3:

\(\begin{array}{l} a,\ \%m_{Zn}=65\%\\ \%m_{Cu}=35\%\\ b,\ m_{\text{dd H$_2$SO$_4$}}=49\ g.\\ c,\ C\%_{\text{dd spư}}=29,11\%\end{array}\)

Giải thích các bước giải:

Câu 1:

\(\begin{array}{l} a,\\ PTHH:2Al+6HCl\to 2AlCl_3+3H_2↑\\ n_{Al}=\dfrac{5,4}{27}=0,2\ mol.\\ Theo\ pt:\ n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\ mol.\\ \Rightarrow V_{H_2}=0,3\times 22,4=6,72\ lít.\\ b,\\ Theo\ pt:\ n_{AlCl_3}=n_{Al}=0,2\ mol.\\ \Rightarrow m_{AlCl_3}=0,2\times 133,5=26,7\ g.\\ c,\\ Theo\ pt:\ n_{HCl}=3n_{Al}=0,6\ mol.\\ \Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,2}=3\ M.\\ d,\\ V_{\text{dd spư}}=V_{\text{dd tpư}}=200\ ml=0,2\ lít.\\ \Rightarrow C_{M_{AlCl_3}}=\dfrac{0,2}{0,2}=1\ M.\end{array}\)

Câu 2:

\(\begin{array}{l} a,\\ PTHH:Zn+H_2SO_4\to ZnSO_4+H_2↑\\ n_{H_2SO_4}=\dfrac{200\times 9,8\%}{98}=0,2\ mol.\\ Theo\ pt:\ n_{H_2}=n_{H_2SO_4}=0,2\ mol.\\ \Rightarrow V_{H_2}=0,2\times 22,4=4,48\ lít.\\ b,\\ Theo\ pt:\ n_{Zn}=n_{H_2SO_4}=0,2\ mol.\\ \Rightarrow m_{\text{dd spư}}=m_{Zn}+m_{\text{dd H$_2$SO$_4$}}-m_{H_2}\\ \Rightarrow m_{\text{dd spư}}=0,2\times 65+200-0,2\times 2=212,6\ g.\\ Theo\ pt:\ n_{ZnSO_4}=n_{H_2SO_4}=0,2\ mol.\\ \Rightarrow C\%_{\text{dd spư}}=\dfrac{0,2\times 161}{212,6}\times 100\%=15,15\%\end{array}\)

Câu 3:

\(\begin{array}{l} \text{Dung dịch H$_2$SO$_4$ có nồng độ 20%}\\ \Rightarrow \text{Dung dịch H$_2$SO$_4$ loãng}\\ a,\\ PTHH:Zn+H_2SO_4\to ZnSO_4+H_2↑\\ \text{(Cu không phản ứng với H$_2$SO$_4$ loãng)}\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\ mol.\\ Theo\ pt:\ n_{Zn}=n_{H_2}=0,1\ mol.\\ \Rightarrow m_{Zn}=0,1\times 65=6,5\ g.\\ \Rightarrow \%m_{Zn}=\dfrac{6,5}{10}\times 100\%=65\%\\ \Rightarrow \%m_{Cu}=100\%-65\%=35\%\\ b,\\ Theo\ pt:\ n_{H_2SO_4}=n_{H_2}=0,1\ mol.\\ \Rightarrow m_{\text{dd H$_2$SO$_4$}}=\dfrac{0,1\times 98}{20\%}=49\ g.\\ c,\\ m_{\text{dd spư}}=m_{Zn}+m_{\text{dd H$_2$SO$_4$}}-m_{H_2}\\ \Rightarrow m_{\text{dd spư}}=6,5+49-0,1\times 2=55,3\ g.\\ Theo\ pt:\ n_{ZnSO_4}=n_{H_2}=0,1\ mol.\\ \Rightarrow C\%_{\text{dd spư}}=\dfrac{0,1\times 161}{55,3}\times 100\%=29,11\%\end{array}\)

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