Question

In the reaction, 2 Al (s) + 6HCL (aq) --> 2ALCL3 (aq) + 3 H2 (g), 2.00 g of Al will react with how many milliliters of 0.500 M HCl?

Answer 1

Answer:

The volume of HCl required is $V = 420 mL$

Explanation:

From the question we are told that

    The chemical equation for this reaction is

             $2 Al _{(s)} + 6 HCl _{(aq)} -----> 2 Al Cl_{3} _{(aq)} + 3 H_2 _{(g)}$

    The mass of Al  is $m__{Al}} = 2.00g$

    The concentration of HCl is $C__{HCl}} = 0.500M$

The number of moles of $Al$ given is $n__{Al}} = \frac{mass \ of Al}{Molar \ mass \ of \ Al}$

   The molar mass of Al is  $M = 27 g/mol$

       So

               $n__{Al} = \frac{2}{27}$

               $n__{Al} = \ 0.07 \ moles$

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

                $x = \frac{0.07 *6}{2}$

                $x = 0.21 \ moles$

Now the volume of HCl can be obtained as

           $Volume(V) = \frac{moles}{concentration }$

 So       $V = \frac{0.21}{0.500}$

             $V = 420 mL$