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:

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

chúc bạn học tốt!

`nH2 = 2.24 / 22.4 = 0.1 ( mol )`

`Mg + H2SO4 -> MgSO4 + H2`

 `0.1       -        0.1`

`MgO + H2SO4 -> MgSO4 + H2`

`mMg = 0.1 * 24 = 2.4 (g)`

`%mMg = $\frac{2.4}{16}$ `* 100% = 15 %` 

`%mMgO = 100% - 15 % = 85%`

nMgO = $\frac{16 - 2.4}{40}$ `= 0.34 ( mol)` 

b) Ta có : `nH2SO4 = 0.34 + 0.1 = 0.44 (mol)`

`V_(H2SO4) = 0.44 / 0.1 = 4.4 ( l )`

c) `Cm_(MgSO4) =` $\frac{0.1 + 0.34}{4.4}$ `= 0.1 M`