In the following reaction: mg + 2hcl → mgcl2 + h2 how many liters of h2 would be produced if you started with 24.3 g of mg?

Equation is as follow,

                                       Mg  + 2 HCl → MgCl₂  +  H₂

According to equation 24.3 g of Mg when reacted with excess of HCl produces 22.4 L (1 mole) of H₂ gas, this statement can be written as,

                          24.3 g of Mg   =   22.4 L of H₂
So,
If 24.3 g of Mg is reacted then X L of H₂ will produce,
Again
                         24.3 g of Mg   =   X L of H₂

Solving for X,

                             X   =  (24.3 g ×22.4 L) ÷ 24.3 g

                             X   =  22.4 L of H₂

Interesting! It means we are provided with exactly one mole of Mg, so obviously one mole of H₂ gas is produced which occupies 22.4 L of volume.