\(2Mg+O2-->2MgO\)(1)
\(MgO+2HCl--.MgCl2+H2O\left(2\right)\)
\(Mg+2HCl--,MgCl2+H2\)
\(n_{Mg}=\frac{2,4}{24}=0,1\left(mol\right)\)
\(n_{O2}=\frac{0,896}{22,4}=0,04\left(mol\right)\)
Lập tỉ lệ
\(n_{Mg}\left(\frac{0,1}{1}\right)>n_{O2}\left(\frac{0,04}{1}\right)=>Mgdư\)
Chất rắn sau pư gồm Mg dư và MgO
\(n_{Mg}=n_{MgO}=2n_{O2}=0,08\left(mol\right)\)
\(n_{Mg}dư=0,1-0,08=0,02\left(mol\right)\)
\(n_{H2}=n_{Mg}dư=0,02\left(mol\right)\)
\(V=V_{H2}=0,02.22,4=0,448\left(l\right)\)
\(n_{HCl}=2n_{Mg}=0,04\left(mol\right)\)
\(n_{HCl\left(2\right)}=2n_{MgO}=0,08\left(mol\right)\)
\(\sum n_{HCl}=0,08+0,04=0,12\left(mol\right)\)
\(a=C_{M\left(HCl\right)}=\frac{0,12}{0,15}=0,8\left(M\right)\)
