Đáp án:

Giải thích các bước giải:

\(1/\\ a,\ PTHH:Fe+2HCl\to FeCl_2+H_2↑\\ n_{Fe}=\dfrac{8,4}{56}=0,15\ mol.\\ Theo\ pt:\ n_{H_2}=n_{Fe}=0,15\ mol.\\ ⇒V_{H_2}=0,15\times 22,4=3,36\ lít.\\ n_{HCl}=2n_{Fe}=0,3\ mol.\\ ⇒m_{HCl}=0,3\times 36,5=10,95\ g.\\ ⇒m_{\text{dd HCl}}=\dfrac{10,95}{10,95\%}=100\ g.\\ c,\ Theo\ pt:\ n_{FeCl_2}=n_{Fe}=0,15\ mol.\\ m_{\text{dd spư}}=m_{Fe}+m_{\text{dd HCl}}-m_{H_2}\\ ⇒m_{\text{dd spư}}=8,4+100-0,15\times 2=108,1\ g.\\ ⇒C\%_{FeCl_2}=\dfrac{0,15\times 127}{108,1}\times 100\%=17,62\%\\ 2/\\ a,\ PTHH:\\ Ca+2H_2O\to Ca(OH)_2+H_2↑\ (1)\\ CaO+H_2O\to Ca(OH)_2\ (2)\\ n_{H_2}=\dfrac{1,68}{22,4}=0,075\ mol.\\ Theo\ pt\ (1):\ n_{Ca}=n_{H_2}=0,075\ mol.\\ ⇒m_{Ca}=0,075\times 40=3\ g.\\ ⇒m_{CaO}=8,6-3=5,6\ g.\\ 3/\\ a,\ PTHH:4Al+3O_2\xrightarrow{t^o} 2Al_2O_3\\ b,\ m_{tăng}=m_{O_2}=89,6\ g.\\ \text{Áp dụng ĐLBT khối lượng ta có:}\\ m_{Al_2O_3}=m_{Al}+m_{O_2}=10,8+9,6=20,4\ g.\)

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