a) \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,3<-----0,6<----0,3<----0,3
=> mMg = 0,3.24 = 7,2 (g)
b) \(C_{M\left(HCl\right)}=\dfrac{0,6}{0,25}=2,4M\)
Mg+2HCl->MgCl2+H2
0,3-----0,6------------0,3
n H2=0,3 mol
=>m Mg=0,3.24=7,2g
CM HCl=\(\dfrac{0,6}{0,25}=2,4M\)