Question

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

Answer 1

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.

$\frac{1 mole}{22.4 L} =\frac{? moles}{8.0 L}$                                <----- Proportion

$8.0 = 22.4(? moles)$                         <----- Cross multiply

$0.36 = ? moles$                                 <----- 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