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8.0 liters I2 forms at STP. How many moles of KI were required for the reaction?

80 Liters I2 Forms At STP How Many Moles Of KI Were Required For The Reaction class=

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Answer:

0.71 moles KI

Explanation:

At STP, there should be 1.0 moles at 22.4 L. However, there are only 8.0 L of I₂. Therefore, we must set up a proportion to find the correct moles of I₂.

[tex]\frac{1 mole}{22.4 L} =\frac{? moles}{8.0 L}[/tex]                                <----- Proportion

[tex]8.0 = 22.4(? moles)[/tex]                         <----- Cross multiply

[tex]0.36 = ? moles[/tex]                                 <----- Divide both sides by 22.4

To find the moles of KI, you can use the mole-to-mole ratio from the balanced equation coefficients. The final answer should have 2 sig figs.

2 KI(aq) + Cl₂(g) ---> 2KCl(aq) + 1 I₂(g)
^                                                   ^

0.36 moles I₂          2 moles KI
----------------------  x  --------------------  =  0.71 moles KI
                                   1 mole I₂

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