Answer:
The volume of HCl required is [tex]V = 420 mL[/tex]
Explanation:
From the question we are told that
The chemical equation for this reaction is
[tex]2 Al _{(s)} + 6 HCl _{(aq)} -----> 2 Al Cl_{3} _{(aq)} + 3 H_2 _{(g)}[/tex]
The mass of Al is [tex]m__{Al}} = 2.00g[/tex]
The concentration of HCl is [tex]C__{HCl}} = 0.500M[/tex]
The number of moles of [tex]Al[/tex] given is [tex]n__{Al}} = \frac{mass \ of Al}{Molar \ mass \ of \ Al}[/tex]
The molar mass of Al is [tex]M = 27 g/mol[/tex]
So
[tex]n__{Al} = \frac{2}{27}[/tex]
[tex]n__{Al} = \ 0.07 \ moles[/tex]
From the balanced equation
2 moles of Al reacts with 6 moles of HCl
So 0.07 moles of Al will react with x
Therefore
[tex]x = \frac{0.07 *6}{2}[/tex]
[tex]x = 0.21 \ moles[/tex]
Now the volume of HCl can be obtained as
[tex]Volume(V) = \frac{moles}{concentration }[/tex]
So [tex]V = \frac{0.21}{0.500}[/tex]
[tex]V = 420 mL[/tex]
