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 → MgCl ₂ + H₂

According to equation 24.3 g of Mg when reacted with excess of HCl produces 22.4 L (1 mole) of H ₂ 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.