Answer
$C_M$ = 1.255 M
Work Step by Step
The reaction that takes place is: Mg + 2HCl = $MgCl_2$ + $H_2$
The atomic weight for Mg is equal to 24 g/mole
24 g Mg..............2 moles HCl
4.47 g Mg...........x moles HCl
=> x = $\frac{4.47*2}{24}$ = 0.3725 moles HCl (consumed)
$C_M$ = $\frac{n}{V}$
$n_{HCl}$ = 2*0.5 = 1 mole (initially)
$n_{HCl final}$ = 1 - 0.3725 = 0.6275 moles
$C_{M final}$ = $\frac{0.6275}{0.5}$ = 1.255 M
