n Mg=\(\dfrac{7,2}{24}\)=0,3 mol
Mg+2HCl->MgCl2+H2
0,3-----0,6 mol
->VHCl=0,1l
=>CmHCl=\(\dfrac{0,6}{0,1}\)=6M
VHCl = 120/1,2 = 100 (ml) = 0,1 (l)
nMg = 7,2/24 = 0,3 (mol)
PTHH: Mg + 2HCl -> MgCl2 + H2
Mol: 0,3 ----> 0,6
CmddHCL = 0,6/0,1 = 6M
\(V_{HCl}=\dfrac{m}{D}=\dfrac{120}{1,2}=100ml=0,1l\)
\(n_{Mg}=\dfrac{7,2}{24}=0,3mol\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
0,3 0,3
\(C_{M_{HCL}}=\dfrac{n}{V}=\dfrac{0,3}{0,1}=3M\)