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

Step-by-step explanation:

Answer 2

0.71 moles KI

Step-by-step 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.

`(1 mole)/(22.4 L) =(? 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