Question

What volume of 6.0 M HCl is required to exactly react with 0.040g Mg?

Answer 1

We need, (i) a stoichiometrically balanced equation........

`Mg(s) + 2HCl(aq) rarr MgCl_2(aq) + H_2(g)uarr`

Explanation:

Given the equation, we need (ii) `"2 equiv"` of acid per `"equiv"` metal.

And thus, `(0.040*g)/(24.31*g*mol^-1)xx2xx(10^3*mL*L^-1)/(6.0*mol*L^-1)=0.55*mL`

This is not a large volume in that the acid is quite concentrated, and the metal is quite low atomic mass.