Đáp án:
\(\begin{array}{l} a,\ m_{\text{dd H$_2$SO$_4$}}=100\ g.\\ b,\ V_{H_2}=2,24\ lít.\\ c,\ C\%_{\text{dd spư}}=C\%_{Al_2(SO_4)_3}=11,22\%\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l} a,\\ PTHH:Zn+H_2SO_4\to ZnSO_4+H_2↑\\ n_{Zn}=\dfrac{6,5}{65}=0,1\ mol.\\ Theo\ pt:\ n_{H_2SO_4}=n_{Zn}=0,1\ mol.\\ \Rightarrow m_{\text{dd H$_2$SO$_4$}}=\dfrac{0,1\times 98}{9,8\%}=100\ g.\\ b,\\ Theo\ pt:\ n_{H_2}=n_{Zn}=0,1\ mol.\\ \Rightarrow V_{H_2}=0,1\times 22,4=2,24\ lít.\\ c,\\ PTHH:2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2↑\\ n_{Al}=\dfrac{5,4}{27}=0,2\ mol.\\ \text{Lập tỉ lệ}\ n_{Al}:n_{H_2SO_4}=\dfrac{0,2}{2}>\dfrac{0,1}{3}\\ \Rightarrow \text{Al dư.}\\ Theo\ pt:\ n_{H_2}=n_{H_2SO_4}=0,1\ mol.\\ \Rightarrow m_{\text{dd spư}}=m_{Al\text{(pư)}}+m_{\text{dd H$_2$SO$_4$}}-m_{H_2}=\dfrac{0,1\times 2}{3}\times 27+100-0,1\times 2=101,6\ g.\\ Theo\ pt:\ n_{Al_2(SO_4)_3}=\dfrac{1}{3}n_{H_2SO_4}=\dfrac{1}{30}\ mol.\\ \Rightarrow C\%_{\text{dd spư}}=C\%_{Al_2(SO_4)_3}=\dfrac{\dfrac{1}{30}\times 342}{101,6}\times 100\%=11,22\%\end{array}\)
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