Given 2AL + 6HCL → 2ALCL3 + 3H2, how many grams of aluminum do I need to produce 11 L of hydrogen gas at STP?

Mass of Aluminum= 8.829 g

Given

Reaction

2Al + 6HCl → 2AlCl₃ + 3H₂

Required

mass of Aluminum

Solution

At STP, 1 mol gas = 22.4 L

For 11 L of Hydrogen :

= 11 : 22.4

= 0.491

From equation, mol ratio Al : H₂ = 2 : 3, so mol Al :

= 2/3 x mol H₂

= 2/3 x 0.491

= 0.327

Mass Aluminum(Ar=27 g/mol) :

= 0.327 mol x 27 g/mol

= 8.829 g

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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