1,PTHH:
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
2,
\(nMg=\dfrac{4,8}{24}=0,2mol\)
có pt: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,2--->0,4----->0,2-------->0,2
\(mMgCl_2=0,2.95=19g\)
\(VH_{2\left(đktc\right)}=0,2.22,4=4,48lit\)
a) Zn + 2HCl --> ZnCl2 + H2
b) Theo ĐLBTKL: mZn + mHCl = mZnCl2 + mH2
=> mH2 = 5,2 + 5,84 - 10,88 = 0,16 (g)