Đáp án:
\(4/\\ b,\ m_{AlCl_3}=26,7\ g.\\ c,\ V_{H_2}=6,72\ lít.\\ 5/\\ b,\ V_{O_2}=3,92\ lít.\\ c,\ \%m_{Al}=81,82\%\\ \%m_{Mg}=18,18\%\\ 6/\\ \text{b, Zn dư, $m_{Zn}(dư)=1,95\ g.$}\\ c,\ V_{H_2}=2,688\ lít.\)
Giải thích các bước giải:
\(4/\\ a,\ PTHH:2Al+6HCl\to 2AlCl_3+3H_2↑\\ b,\ n_{Al}=\dfrac{5,4}{27}=0,2\ mol.\\ Theo\ pt:\ n_{AlCl_3}=n_{Al}=0,2\ mol.\\ ⇒m_{AlCl_3}=0,2\times 133,5=26,7\ g.\\ c,\ Theo\ pt:\ n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\ mol.\\ ⇒V_{H_2}=0,3\times 22,4=6,72\ lít.\\ 5/\\ a,\ PTHH:\\ 2Mg+O_2\xrightarrow{t^o} 2MgO\ (1)\\ 4Al+3O_2\xrightarrow{t^o} 2Al_2O_3\ (2)\\ \text{b, Áp dụng ĐLBT khối lượng ta có:}\\ m_{O_2}=m_{\text{oxit}}-m_{\text{hỗn hợp}}=12,2-6,6=5,6\ g.\\ ⇒n_{O_2}=\dfrac{5,6}{32}=0,175\ mol.\\ ⇒V_{O_2}=0,175\times 22,4=3,92\ lít.\\ c,\ \text{Gọi $n_{Al}$ là a (mol), $n_{Mg}$ là b (mol).}\\ \text{Theo đề bài ta có hệ pt:}\\ \left\{\begin{matrix} 27a+24b=6,6 & \\ 0,75a+0,5b=0,175 & \end{matrix}\right.\\ ⇒\left\{\begin{matrix} a=0,2 & \\ b=0,05 & \end{matrix}\right.\\ ⇒\%m_{Al}=\dfrac{0,2\times 27}{6,6}\times 100\%=81,82\%\\ ⇒\%m_{Mg}=\dfrac{0,05\times 24}{6,6}\times 100\%=18,18\%\\ 6/\\ a,\ PTHH:Zn+2HCl\to ZnCl_2+H_2↑\\ b,n_{Zn}=\dfrac{9,75}{65}=0,15\ mol.\\ n_{HCl}=\dfrac{8,76}{36,5}=0,24\ mol.\\ \text{Lập tỉ lệ:}\ \dfrac{0,15}{1}>\dfrac{0,24}{2}\\ ⇒Zn\ dư.\\ ⇒n_{Zn}(dư)=0,15-\dfrac{0,24}{2}=0,03\ mol.\\ ⇒m_{Zn}(dư)=0,03\times 65=1,95\ g.\\ c,\ Theo\ pt:\ n_{H_2}=\dfrac{1}{2}n_{HCl}=0,12\ mol.\\ ⇒V_{H_2}=0,12\times 22,4=2,688\ lít.\)
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