potassium iodate and potassium iodite react with sulfuric acid to produce molecular iodine in excess
KIO3 + 5KI + 6H+ =>3I2 + 6K+ + 3H2O
If I add 0.268 grams of KIO3 to 5 grams of KI and make it up to a 200 ml solution (with distilled water); then add:
30 ml of H2SO4 (provides 6H+) and then make it up to a 500 ml solution (with distilled water).
What concentration of Iodine do I get (what concentration solution is this?) and how many moles of Iodine do I get as a product.
i understand mole calculations, but i have never come across one like this; i do not understand how to work out the moles of iodine i get when i have added different masses of KIO and KI.
Do you get what i mean? i don't understand how to put the two mole values together.
Please help, and explain in a simple way so that i can understand each step involved.
Thank you!
:'(
KIO3 + 5KI + 6H+ =>3I2 + 6K+ + 3H2O
If I add 0.268 grams of KIO3 to 5 grams of KI and make it up to a 200 ml solution (with distilled water); then add:
30 ml of H2SO4 (provides 6H+) and then make it up to a 500 ml solution (with distilled water).
What concentration of Iodine do I get (what concentration solution is this?) and how many moles of Iodine do I get as a product.
i understand mole calculations, but i have never come across one like this; i do not understand how to work out the moles of iodine i get when i have added different masses of KIO and KI.
Do you get what i mean? i don't understand how to put the two mole values together.
Please help, and explain in a simple way so that i can understand each step involved.
Thank you!
:'(
-
First calculate how many moles of the two reactants you have.
0.268 g KIO3 x (1 mole KIO3 / 214.0 g KIO3) = 0.00125 moles KIO3
5.00 g KI x (1 mole KI / 166.0 g KI) = 0.0301 moles KI
The balanced equation tells us that it takes 5 moles of KI to react with 1 mole of KIO3.
0.00125 moles KIO3 x (5 moles KI / 1 mole KIO3) = 0.00625 moles KI needed to react
Obviously we have plenty of KI (0.0301 moles), in fact, a large excess. So the moles of KIO3 will determine how much I2 we can make. The balanced equation tells us that 1 mole of KIO3 produces 3 moles of I2.
0.00125 moles KIO3 x (3 moles I2 / 1 mole KIO3) = 0.00375 moles I2 produced.
The molarity of I2 in the final solution is
M I2 = moles I2 / L of solution = 0.00375 / 0.500 = 0.00750 M I2
0.268 g KIO3 x (1 mole KIO3 / 214.0 g KIO3) = 0.00125 moles KIO3
5.00 g KI x (1 mole KI / 166.0 g KI) = 0.0301 moles KI
The balanced equation tells us that it takes 5 moles of KI to react with 1 mole of KIO3.
0.00125 moles KIO3 x (5 moles KI / 1 mole KIO3) = 0.00625 moles KI needed to react
Obviously we have plenty of KI (0.0301 moles), in fact, a large excess. So the moles of KIO3 will determine how much I2 we can make. The balanced equation tells us that 1 mole of KIO3 produces 3 moles of I2.
0.00125 moles KIO3 x (3 moles I2 / 1 mole KIO3) = 0.00375 moles I2 produced.
The molarity of I2 in the final solution is
M I2 = moles I2 / L of solution = 0.00375 / 0.500 = 0.00750 M I2