Question

Is this the right answer that I have?

What is the minimum volume of 2.49 mol L−1 HCl(aq) required to dissolve 10.5 g Al metal?

The atomic weight of Al is 26.98 g mol−1.

Al(s) + HCl(aq) ⟶ AlCl3(aq) + H2(g) (unbalanced)

would it be 0.469? Am i right here?

Answer 1

0.469 L

Step-by-step explanation:

In order to find the minimum volume of 2.49 mol HCl(aq) required to dissolve 10.5 g of Al metal, we must first balance the chemical equation:

2Al(s) + 6HCl(aq) ⟶ 2AlCl3(aq) + 3H2(g)

We do this by making sure the amount of each element is equal on both sides of the equation.

We can then determine the amount of moles in 10.5 g of Al. We do this by dividing the mass by the molar mass:

10.5 g Al ÷ 26.98 g/mol = 0.389 mol Al

Now we can determine how many of moles of HCl(aq) will react:

0.389 mol Al ×
`(6 mol HCl)/(2 mol Al)` = 1.17 mol HCl

Now, we can finally find the volume of 2.49 mol/L HCl required for the reaction. We do this by dividing the amount of moles by the concentration:

1.17 mol HCl ÷ 2.49 mol/L = 0.469 L

You are correct!