$a,PTHH:MgO+2HCl\to MgCl_2+H_2O\\ Fe+2HCl\to FeCl_2+H_2\\ b,n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)\\ \Rightarrow n_{Fe}=n_{H_2}=0,2(mol)\\ \Rightarrow m_{Fe}=0,2.56=11,2(g)\\ \Rightarrow m_{MgO}=15,2-11,2=4(g)\\ c,n_{MgO}=\dfrac{4}{40}=0,1(mol)\\ \Rightarrow \Sigma n_{HCl}=2n_{MgO}+2n_{Fe}=0,4+0,2=0,6(mol)\\ \Rightarrow m_{HCl}=0,6.36,5=21,9(g)$

Em tham khảo!

a) Ta có PTHH đầy đủ sau:

$\text{1) MgO+2HCl$\rightarrow$MgCl2+H2O}$

$\text{2) Fe+2HCl$\rightarrow$FeCl2+H2}$

b)

Có $\text{nH2=$\dfrac{4,48}{22,4}$=0,2 mol}$

Dựa vào PTHH 2) $\text{nH2=nFe=0,2 mol}$

Vậy $\text{mFe=0,2.56=11,2g}$

Vậy $\text{mMgO=15,2-11,2=4g}$

c)

Có $\text{nHCl=2nH2=0,2.2=0,4 mol}$

Có $\text{nMgO=$\dfrac{4}{40}$=0,1 mol}$

Dựa vào PTHH 1) $\text{nHCl=2nMgO=0,1.2=0,2 mol}$

Vậy $\text{∑nHCl=0,4+0,2=0,6 mol}$

$\rightarrow$ $\text{mHCl=0,6.36,5=21,9g}$