Question

What is the mass in grams of Al that were reacted with excess HCl if 2.12 L of hydrogen gas were collected at STP in the following reaction?
2 Al (s) + 6 HCl (aq) → 2 AlCl₃ (aq) + 3 H₂ (g)

Answer 1

4.176 g of Al that was reacted with excess HCl if 2.12 L of hydrogen gas were collected at STP in the following reaction.

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

What is an ideal gas equation?

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

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

Given data:

Volume of $H_2$ (g) = 5.20L

At STP

Pressure= 1 atm

Temperature =273K

R = 0.0821 L.atm/mol K

PV=nRT

n= $\frac{PV}{RT}$

n= $\frac{1 atm X 5.20 L}{0.0821L.atm/mol K}$

n= 0.2324 mol $H_2$

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

=$0.2324 mol H_2 X$ $\frac{2 mol Al}{3 mol H_2} X\frac{27 g Al}{1 mol Al}$

=4.176 g of Al

Hence, 4.176 g of Al that were reacted with excess HCl if 2.12 L of hydrogen gas were collected at STP in the following reaction.

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Answer 2

The mass of Al that reacted with excess to produce 2.12 L of Hydrogen gas at STP will be 1.7 gm

What is Limiting reagent ?

The limiting reagent  is the reactant that gets consumed first in a chemical reaction and therefore limits how much product can be formed.

We're asked to find the number of grams of Al that reacted with excess to produce 2.12 L of Hydrogen gas at STP

Balanced chemical equation for this reaction :

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

From the above equation, Using Limiting reagent concept, we can conclude that 3 mole of Hydrogen gas i.e, 67.2 Ltr  (3 mole x 22.4 ltr) Hydrogen gas is produced by the consumption of 2 Al i.e, 54 gm (Mass = 2 mole x 27 g)

Therefore,

Mass of Al required to produce 2.12 ltr of Hydrogen gas  = 54/67.2 x 2.12 = 1.7 gm

Hence, the mass of Al that reacted with excess to produce 2.12 L of Hydrogen gas at STP will be 1.7 gm

9

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