\(n_{Mg}=0,25\left(mol\right)\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
0,25 0,5 0,25 0,25
a ) \(D=\dfrac{m_{dd\left(HCl\right)}}{V}\Rightarrow m_{dd\left(HCl\right)}=50.1,2=60\left(g\right)\)
\(m_{HCl}=0,5.35,5=17,75\left(g\right)\)
\(C\%_{\left(HCl\right)}=\dfrac{m_{ct}}{m_đ}=\dfrac{17,75}{60}.100\%=29,5\%\)
b ) \(m_{dd\left(sau\right)}=m_{Mg}+m_{dd\left(HCl\right)}-m_{H_2\uparrow}=6+50-0,5=55,5\)
\(m_{MgCl_2}=0,25.95=23,75\)
\(C\%_{\left(MgCl_2\right)}=\dfrac{m_{ct}}{m_{dd}}=\dfrac{23,75}{55,5}.100\%=42,79\%\)
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