Question

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)

Answer 1

The volume in liters of Hydrogen gas = 49.28 L

Further explanation

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 :

$\tt mol=\dfrac{mass}{Ar}=\dfrac{40}{27}=1.481$

mol  H₂ :

$\tt \dfrac{3}{2}\times 1.481=2.2$

At STP, 1 mol =22.4 L

so the volume of hydrogen :

$\tt 2.2 \times 22.4=49.28~L$

Answer 2

Answer:

49.8 L

Explanation:

40.0 g Al x 1 mol Al / 26.98g Al x 3 mol H2 / 2 mol Al = 2.22 mol H2

PV = nRT

V = nRT / P

(2.22mol x 0.08206 x 273.15 K) / 1 atm = 49.8 L