Đáp án:
\(\begin{array}{l} a,\ \%m_{Mg}=15\%\\ \%m_{MgO}=85\%\\ b,\ V_{\text{dd H$_2$SO$_4$}}=4,4\ lít.\\ c,\ C_{M_{MgSO_4}}=0,1\ M.\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l} a,\\ PTHH:\\ Mg+H_2SO_4\to MgSO_4+H_2↑\ (1)\\ MgO+H_2SO_4\to MgSO_4+H_2O\ (2)\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\ mol.\\ Theo\ pt\ (1):\ n_{Mg}=n_{H_2}=0,1\ mol.\\ \Rightarrow \%m_{Mg}=\dfrac{0,1\times 24}{16}\times 100\%=15\%\\ \Rightarrow \%m_{MgO}=100\%-15\%=85\%\\ b,\\ m_{MgO}=16\times 85\%=13,6\ g.\\ \Rightarrow n_{MgO}=\dfrac{13,6}{40}=0,34\ mol.\\ \Rightarrow ∑n_{H_2SO_4}=n_{Mg}+n_{MgO}=0,1+0,34=0,44\ mol.\\ \Rightarrow V_{\text{dd H$_2$SO$_4$}}=\dfrac{0,44}{0,1}=4,4\ lít.\\ c,\\ ∑n_{MgSO_4}=n_{Mg}+n_{MgO}=0,1+0,34=0,44\ mol.\\ V_{\text{dd spư}}=V_{\text{dd tpư}}=4,4\ lít.\\ \Rightarrow C_{M_{MgSO_4}}=\dfrac{0,44}{4,4}=0,1\ M.\end{array}\)
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