Đáp án:

\(\begin{array}{l} a,\ m_{Mg}=4,8\ g.\\ b,\ C\%_{\text{dd HCl}}=7,3\%\\ c,\ C\%_{\text{dd spư}}=C\%_{MgCl_2}=9,3\%\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l} a,\\ PTHH:Mg+2HCl\to MgCl_2+H_2↑\\ n_{H_2}=\dfrac{4,48}{22,4}=0,2\ mol.\\ Theo\ pt:\ n_{Mg}=n_{H_2}=0,2\ mol.\\ \Rightarrow m_{Mg}=0,2\times 24=4,8\ g.\\ b,\\ Theo\ pt:\ n_{HCl}=2n_{H_2}=0,4\ mol.\\ \Rightarrow C\%_{\text{dd HCl}}=\dfrac{0,4\times 36,5}{200}\times 100\%=7,3\%\\ c,\\ m_{\text{dd spư}}=m_{Mg}+m_{\text{dd HCl}}-m_{H_2}\\ \Rightarrow m_{\text{dd spư}}=4,8+200-0,2\times 2=204,4\ g.\\ Theo\ pt:\ n_{MgCl_2}=n_{H_2}=0,2\ mol.\\ \Rightarrow C\%_{\text{dd spư}}=C\%_{MgCl_2}=\dfrac{0,2\times 95}{204,4}\times 100\%=9,3\%\end{array}\)

chúc bạn học tốt!

a,

$Mg+2HCl\to MgCl_2+H_2$

$n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)$

$\Rightarrow n_{Mg}=0,2(mol)$

$m_{Mg}=0,2.24=4,8g$

b,

$n_{HCl}=2n_{H_2}=0,2.2=0,4(mol)$

$\Rightarrow C\%_{HCl}=\dfrac{0,4.36,5.100}{200}=7,3\%$

c,

$m_{dd\text{spứ}}=4,8+200-0,2.2=204,4g$

$n_{MgCl_2}=n_{Mg}=0,2(mol)$

$\Rightarrow C\%_{MgCl_2}=\dfrac{0,2.95.100}{204,4}=9,3\%$