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Vladimir [108]
3 years ago
13

The equilibrium constant for the reaction H2 (g) + I2 (g) <==> 2 HI (g)

Chemistry
1 answer:
antoniya [11.8K]3 years ago
5 0

The percentage of I2 converted = 78.6 %

Explanation:

Write down the given values

Kp= 54.4

The percentage of I2 will be converted to HI is 0.200 moled each of H2 and I2.

It should come to an equilibrium in 1.00 lit container .

Change in I2 (iodine) and H2 (hydrogen)  = x in each

Change in HI = 2x

total ni (nickel)

number of moles = 0.2 -x + 0.2 -x + 2x

=0.4 moles

Mole fractions :

I2 = 0.2-x / 0.4 H2

=0.2-x  / 0.4 HI

= 2x /0.4

Kp = HI ^2 / H2* I2

= (2x) ^2 / (0.2-x) ^2 = 54.4

by  taking square root:

2x / 0.2-x = 7.375

= x=0.157

percentage of I2 converted = 78.6 %

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how many moles of iron will be produced from 6.20 moles of carbon monoxide reacting with excess iron (III) oxide (FeO3) to produ
Gelneren [198K]

0.20 moles of iron will be formed in the reaction.

Explanation:

The balanced chemical equation for the reaction between iron (iii) oxide and carbon monoxide to form Fe is to be known first.

the balanced reaction is :

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so from the data given the number of moles of carbon monoxide can be known:

3 moles of CO reacted with Fe2O3 to form 2 moles of iron in the reaction.

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5 0
3 years ago
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8 0
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