a.\(n_{Mg}=\dfrac{4,8}{24}=0,2mol\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,2 0,2 ( mol )
\(V_{H_2}=0,2.24,79=4,958l\)
b.\(n_{Na}=\dfrac{6,9}{23}=0,3mol\)
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
0,3 0,15 ( mol )
\(V_{H_2}=0,15.24,79=3,7185l\)
c.\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
0,05 0,075 ( mol )
\(V_{H_2}=0,075.24,79=1,85925l\)