Đáp án:
\(\begin{array}{l} a,\ \%m_{Mg}=44,44\%\\ \%m_{MgO}=55,56\%\\ b,\ V_{HCl}=0,35\ lít.\\ c,\ m_{MgCl_2}=33,25\ g.\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l} a,\ PTHH:\\ Mg+2HCl\to MgCl_2+H_2↑\ (1)\\ MgO+2HCl\to MgCl_2+H_2O\ (2)\\ n_{H_2}=\dfrac{4,48}{22,4}=0,2\ mol.\\ Theo\ pt\ (1):\ n_{Mg}=n_{H_2}=0,2\ mol.\\ ⇒m_{Mg}=0,2\times 24=4,8\ g.\\ ⇒m_{MgO}=10,8-4,8=6\ g.\\ ⇒\%m_{Mg}=\dfrac{4,8}{10,8}\times 100\%=44,44\%\\ \%m_{MgO}=\dfrac{6}{10,8}\times 100\%=55,56\%\\ b,\ n_{MgO}=\dfrac{6}{40}=0,15\ mol.\\ ⇒∑n_{HCl}=2n_{H_2}+2n_{MgO}=0,7\ mol.\\ ⇒V_{HCl}=\dfrac{0,7}{2}=0,35\ lít.\\ c,\ ∑n_{MgCl_2}=n_{Mg}+n_{MgO}=0,2+0,15=0,35\ mol.\\ ⇒m_{MgCl_2}=0,35\times 95=33,25\ g.\end{array}\)
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