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 :

ardni313

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 :

[tex]\tt mol=\dfrac{mass}{Ar}=\dfrac{40}{27}=1.481[/tex]

mol  H₂ :

[tex]\tt \dfrac{3}{2}\times 1.481=2.2[/tex]

At STP, 1 mol =22.4 L

so the volume of hydrogen :

[tex]\tt 2.2 \times 22.4=49.28~L[/tex]

dsdrajlin

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.

  • The molar mass of Al is 26.98 g/mol.
  • The molar ratio of Al to H₂ is 2:3.
  • 1 mole of H₂ at STP occupies 22.4 L.

[tex]40.0 g Al \times \frac{1molAl}{26.98gAl} \times \frac{3molH_2}{2molAl} \times \frac{22.4 LH_2}{1molH_2} = 49.8 LH_2[/tex]

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

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