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

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

Step-by-step 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)} = (mass \ of Al)/(Molar \ mass \ of \ Al)`

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

So

`n__(Al) = (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 = (0.07 *6)/(2)`

`x = 0.21 \ moles`

Now the volume of HCl can be obtained as

`Volume(V) = (moles)/(concentration )`

So
`V = (0.21)/(0.500)`

`V = 420 mL`