Đáp án:
\(\begin{array}{l} a,\ m_{Mg}=4,8\ g.\\ m_{MgO}=4\ g.\\ b,\ m_{HCl}=21,9\ g.\\ m_{\text{dd HCl}}=150\ g.\\ c,\ C\%_{MgCl_2}=17,99\%\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l} a,\\ PTHH:Mg+2HCl\to MgCl_2+H_2↑\ (1)\\ MgO+2HCl\to MgCl_2+H_2O\ (2)\\ n_{MgCl_2}=\dfrac{28,5}{95}=0,3\ mol.\\ \text{Gọi $n_{Mg}$ là a (mol), $n_{MgO}$ là b (mol).}\\ \text{Theo đề bài ta có hệ pt:}\ \begin{cases}24a+40b=8,8\\a+b=0,3\end{cases}\ \Rightarrow \begin{cases}a=0,2\\b=0,1\end{cases}\\ \Rightarrow m_{Mg}=0,2\times 24=4,8\ g.\\ m_{MgO}=0,1\times 40=4\ g.\\ b,\\ ∑n_{HCl}=2n_{Mg}+2n_{MgO}=0,4+0,2=0,6\ mol.\\ \Rightarrow m_{HCl}=0,6\times 36,5=21,9\ g.\\ \Rightarrow m_{\text{dd HCl}}=\dfrac{21,6}{14,6\%}=150\ g.\\ c,\\ Theo\ pt\ (1):\ n_{H_2}=n_{Mg}=0,2\ mol.\\ \Rightarrow m_{\text{dd spư}}=m_{A}+m_{\text{dd HCl}}-m_{H_2}\\ \Rightarrow m_{\text{dd spư}}=8,8+150-0,2\times 2=158,4\ g.\\ \Rightarrow C\%_{MgCl_2}=\dfrac{28,5}{158,4}\times 100\%=17,99\%\end{array}\)
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