when 2.0 mol h2 and 2.0 mol i2 are placed in a reaction vessel and allowed to come to equilibrium, the mixture is found to contain 2.5 mol hi. how many moles of each substance are present in the equilibrium mixture, i.e. what is the composition?

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21-10-2022
Chemistry

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when 2.0 mol h2 and 2.0 mol i2 are placed in a reaction vessel and allowed to come to equilibrium, the mixture is found to contain 2.5 mol hi. how many moles of each substance are present in the equilibrium mixture, i.e. what is the composition?

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kjuli1766 kjuli1766 · Answer 1 · 2022-10-25T14:27:35+02:00

The composition of the mixture is 2.5 moles of HI, 0.75-mole I₂, and 0.75 moles of H₂.

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H₂ + I₂ = 2HI

initial 2mol 2 0mol

final 2.5 mol

limiting reagent is I₂

2 mole of HI is produced by 1 mole of I₂

2.5 mole of I₂ is produced by = 1/2 × 2.5

= 1.25 mole

The final composition of the mixture = 2.5 moles of HI, 0.75-mole I₂, and 0.75 moles of H₂.

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