Question

Question #a4750

Answer 1

Your reaction will produce 6.2 L of iodine gas at STP.

SIDE NOTE I'm almost certain that you are supposed to use the 22.4 L value for the molar volume of a gas at STP.

This value is incorrect, as it comes from the old IUPAC definition for STP, which was 1 atm and 0 degrees Celsius. The current condtions for STP are 100 kPa and 0 degrees Celsius, which imply a molar volume equal to 22.7 L.

I'll use the correct value for STP; if you were supposed to use the old value, simply redo all the calculations.

So, the balanced chemical equation for your reaction is

`color(red)(2)KI_((aq)) + Cl_(2(g)) -> 2KCl_((aq)) + I_(2(g))`

Notice that you have a `color(red)(2):1` mole ratio between potassium iodide and chlorine gas, and a `1:1` mole ratio between chlorine gas and iodine gas.

This means that, regardless of how many moles of chlorine gas you have, you need twice more moles of potassium iodide for the reaction to take place.

Using the molar volume of a gas at STP, the number of moles of `Cl_(2)` will be

`n = V/V_"molar" = (22cancel("L"))/(22.7cancel("L")/"mol") = "0.9692 moles "` `Cl_2`

This much chlorine gas would require

`0.9692cancel("moles Cl"_2) * "2 moles KI"/(1cancel("mole Cl"_2)) = "1.9384 moles KI"`

The actual number of moles of potassium iodide available is

`C = n/V => n = C *V`

`n_(KI) = 2.15"mol"/cancel("L") * 255 * 10^(-3)cancel("L") = "0.5483 moles KI"`

This means that potassium iodide will act as a limiting reagent, i.e. it will determine how many moles of chlorine will actually react.

`0.5483cancel("moles KI") * ("1 mole "Cl_2)/(2cancel("moles KI")) ="0.2741 moles "` `Cl_2`

Since this is also the number of moles of iodine gas produced, the volume will be equal to

`V = n * V_"molar" = 0.2741cancel("moles") * 22.7"L"/cancel("mol") = "6.222 L"`

Rounded to two sig figs, the number of sig figs given for 22 L, the answer will be

`V_(I_2) = color(green)("6.2 L")`