Determine the moles of H+ reacting with the metal based on the following experimental data. 10.00 mL of 1.00 M HCl solution is added to a sample of Mg metal. The reaction goes to completion and all of the Mg metal is gone. 27.08 mL of 0.30 M NaOH solution is required to reach the end point when titrating the remaining HCl.

[tex]10.00 \; \rm mL = 10.00 \times 10^{-3}\; \rm L = 1.000 \times 10^{-2}\; \rm L[/tex].
[tex]27.08 \; \rm mL = 27.08 \times 10^{-3}\; \rm L = 2.708 \times 10^{-2}\; \rm L[/tex].

Number of moles of [tex]\rm HCl[/tex] initially present (in the [tex]10.00\; \rm mL[/tex] solution at [tex]1.00\; \rm M[/tex].)

[tex]n(\mathrm{HCl}, \, \text{initial}) = \displaystyle c \cdot V = 1.00\times 1.000 \times 10^{-2}= 1.000\times 10^{-2}\; \rm mol[/tex].

Number of moles of [tex]\rm NaOH[/tex] from the titration:

[tex]n(\mathrm{NaOH}) = c \cdot V = 0.30 \times 2.708 \times 10^{-2} = 8.124 \times 10^{-3}\; \rm mol[/tex].

[tex]\rm NaOH[/tex] neutralizes [tex]\rm HCl[/tex] at a [tex]1:1[/tex] ratio:

[tex]\rm HCl + NaOH \to NaCl + H_2O[/tex].

Hence, [tex]n(\mathrm{HCl},\, \text{leftover}) = n(\mathrm{NaOH}) = 8.124 \times 10^{-3}\; \rm mol[/tex].

[tex]\begin{aligned}&n(\mathrm{HCl},\, \text{consumed}) \\ =& n(\mathrm{HCl},\, \text{initial}) - n(\mathrm{HCl},\, \text{leftover}) \\ =& 1.000\times 10^{-2} - 8.124\times 10^{-3} \\ =& 1.876 \times 10^{-3}\; \rm mol\end{aligned}[/tex].