Question

12grams of magnesium is made to react with enough amount of HCl.what volume of hydrogen gas will be collected at S.T.P? THANKS

Answer 1

We need a stoichiometric equation.....and I make the volume to be a tad over `11*dm^3`

Explanation:

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

We know (i) the molar quantity of the metal, i.e. `(12.0*g)/(24.3*g*mol^-1)=0.494*mol`....

And given the stoichiometric reaction, we know that `0.494*mol` dihydrogen gas are evolved...

We further need (ii) the molar volume of an Ideal Gas at `"STP"`, and this site quotes a value of `22.4*dm^3*mol^-1`...

And so we take the product....`22.4*dm^3*mol^-1xx0.494*mol`

`=11.1*dm^3`

And note that `1*dm^3=(1xx10^-1*m)^3=10^-3*m^3`

`1/1000*m^3=1*L` because there are `10^3*L*m^-3`...