What is the volume in liters of hydrogen gas that would be produced by the reaction of 40.0 g of Al with excess HCl at STP according to the following reaction? 2 Al (s) + 6 HCl (aq) → 2 AlCl₃ (aq) + 3 H₂ (g)

When 40.0 g of Al react with excess HCl at STP, 49.8 L of H₂ are produced.

Let's consider the following balanced equation.

2 Al(s) + 6 HCl(aq) → 2 AlCl₃(aq) + 3 H₂(g)

We can calculate the liters of hydrogen obtained by the reaction of 40.0 g of Al with excess HCl considering the following relationships.

When 40.0 g of Al react with excess HCl at STP, 49.8 L of H₂ are produced.

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The volume in liters of Hydrogen gas = 49.28 L

The reaction equation is the chemical formula of reagents and product substances  

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products  

Reaction

2 Al (s) + 6 HCl (aq) → 2 AlCl₃ (aq) + 3 H₂ (g)

mol Al :

mol  H₂ :

At STP, 1 mol =22.4 L

so the volume of hydrogen :

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