Question

At `"STP"` what mass of aluminum metal is required to give a `0.6225*L` volume of dihydrogen gas?

Answer 1

`"STP"` specifies that the volume of a molar quantity of gas is `22.4*dm^3-=22.4*L`...(depending on the syllabus....)..we need about a half a gram of aluminum....

Explanation:

And so.......... `"moles of dihydrogen"=(0.6225*L)/(22.4*L*mol^-1)=0.0278*mol.`

And we refer to the stoichiometric equation....

`Al(s) + 3HCl(aq) rarr AlCl_3(aq) + 3/2H_2(g)uarr`

And thus we require `2/3*"equiv"` of aluminum metal, i.e. `0.0185*mol`, i.e. a mass of `0.0185*molxx26.98*g*mol^-1=??*g`