Đáp án:
Câu 3: (trên)
\(\begin{array}{l} a,\ \text{X là Nhôm (Al).}\\ b,\ C\%_{\text{HCl (dư)}}=3,435\%\\ C\%_{AlCl_3}=19,33\%\end{array}\)
Câu 3: (dưới)
\(\begin{array}{l} a,\ \text{X là Natri (Na).}\\ b,\ C_{M_{\text{HCl (dư)}}}=1,13\ (M).\\ C_{M_{NaOH}}=0,43\ (M).\end{array}\)
Giải thích các bước giải:
Câu 3: (trên)
\(\begin{array}{l} a,\\ PTHH:\ 2X+6HCl\to 2XCl_3+3H_2\uparrow\\ n_{HCl}=\dfrac{400\times 20\%}{36,5}≈2,19\ (mol).\\ n_{H_2}=\dfrac{20,16}{22,4}=0,9\ (mol).\\ \text{Lập tỉ lệ:}\ n_{HCl}:n_{H_2}=\dfrac{2,19}{2}>\dfrac{0,9}{1}\\ \Rightarrow HCl\ \text{dư.}\\ Theo\ pt:\ n_{X}=\dfrac{2}{3}n_{H_2}=0,6\ (mol).\\ \Rightarrow M_{X}=\dfrac{16,2}{0,6}=27\ (g/mol)\\ \Rightarrow \text{X là Nhôm (Al).}\\ b,\\ PTHH:2Al+6HCl\to 2AlCl_3+3H_2\uparrow\\ n_{HCl\ \text{(dư)}}=2,19-(0,9\times 2)=0,39\ (mol).\\ Theo\ pt:\ n_{AlCl_3}=n_{Al}=0,6\ (mol).\\ m_{\text{dd spư}}=m_{Al}+m_{\text{dd HCl}}-m_{H_2}\\ \Rightarrow m_{\text{dd spư}}=16,2+400-0,9\times 2=414,4\ (g).\\ \Rightarrow C\%_{\text{HCl (dư)}}=\dfrac{0,39\times 36,5}{414,4}\times 100\%=3,435\%\\ C\%_{AlCl_3}=\dfrac{0,6\times 133,5}{414,4}\times 100\%=19,33\%\end{array}\)
Câu 3: (dưới)
\(\begin{array}{l} a,\\ PTHH:2X+2HCl\to 2XCl+H_2\uparrow\\ n_{HCl}=0,3\times 2=0,6\ (mol).\\ n_{H_2}=\dfrac{0,26}{2}=0,13\ (mol).\\ \text{Lập tỉ lệ:}\ n_{HCl}:n_{H_2}=\dfrac{0,6}{2}>\dfrac{0,13}{1}\\ \Rightarrow \text{HCl dư.}\\ Theo\ pt:\ n_{X}=2n_{H_2}=0,26\ (mol).\\ \Rightarrow M_{X}=\dfrac{5,98}{0,26}=23\ (g/mol)\\ \Rightarrow \text{X là Natri (Na).}\\ b,\\ PTHH:2Na+2H_2O\to 2NaOH+H_2\uparrow\\ V_{\text{dd spư}}=V_{\text{dd tpư}}=300\ (ml)=0,3\ (l).\\ n_{HCl\text{(dư)}}=0,6-(0,13\times 2)=0,34\ (mol).\\ Theo\ pt:\ n_{NaOH}=n_{Na}=0,13\ (mol).\\ \Rightarrow C_{M_{\text{HCl (dư)}}}=\dfrac{0,34}{0,3}=1,13\ (M).\\ C_{M_{NaOH}}=\dfrac{0,13}{0,3}=0,43\ (M).\end{array}\)
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