Respuesta :
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
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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 [tex]H_2[/tex] (g) = 5.20L
At STP
Pressure= 1 atm
Temperature =273K
R = 0.0821 L.atm/mol K
PV=nRT
n= [tex]\frac{PV}{RT}[/tex]
n= [tex]\frac{1 atm X 5.20 L}{0.0821L.atm/mol K}[/tex]
n= 0.2324 mol [tex]H_2[/tex]
2 Al (s) + 6 HCl (aq) → 2 AlCl₃ (aq) + 3 H₂ (g)
=[tex]0.2324 mol H_2 X[/tex] [tex]\frac{2 mol Al}{3 mol H_2} X\frac{27 g Al}{1 mol Al}[/tex]
=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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