AS chem

Here is an exam question.

0.50 moles of chlorine gas were passed into an aqueous solution containing 0.66 moles of each of sodium bromide and sodium iodide. Assuming that all of the chlorine reacted, calculate;
(i) the number of moles of iodine produced;
(ii) the number of moles of bromine produced.

Reply 1

The Chameleon

OP

Anyone?

Reply 2

oxymoron

7

Cl2 + 2NaX --> 2NaCl+ X2 (where X = Br or I)

so one mole of Cl2 reacts with 2 moles of NaX

so 0.5 moles of Cl2 react with 1 mol of NaX

so 0.66 mol NaI react (as this reaction is more favourable due to bigger gap in reactivity) leaving 1-0.66=0.34 mol of NaBr

This will produce 0.66/2 = 0.33 mol of I2
and 0.34/2=0.17 mol Br2

Reply 3

The Chameleon

OP

oxymoron

Cl2 + 2NaX --> 2NaCl+ X2 (where X = Br or I)

so one mole of Cl2 reacts with 2 moles of NaX

so 0.5 moles of Cl2 react with 1 mol of NaX

so 0.66 mol NaI react (as this reaction is more favourable due to bigger gap in reactivity) leaving 1-0.66=0.34 mol of NaBr

This will produce 0.66/2 = 0.33 mol of I2
and 0.34/2=0.17 mol Br2

I really don't understand what you've done. I've never had to do this in any calculations before.

Reply 4

supreme

4

oxymoron

Cl2 + 2NaX --> 2NaCl+ X2 (where X = Br or I)

so one mole of Cl2 reacts with 2 moles of NaX

so 0.5 moles of Cl2 react with 1 mol of NaX

so 0.66 mol NaI react (as this reaction is more favourable due to bigger gap in reactivity) leaving 1-0.66=0.34 mol of NaBr

This will produce 0.66/2 = 0.33 mol of I2
and 0.34/2=0.17 mol Br2

But would there be enough Cl left for a second reaction with NaBr?

Reply 5

The Chameleon

OP

oxymoron

Cl2 + 2NaX --> 2NaCl+ X2 (where X = Br or I)

so one mole of Cl2 reacts with 2 moles of NaX

so 0.5 moles of Cl2 react with 1 mol of NaX

so 0.66 mol NaI react (as this reaction is more favourable due to bigger gap in reactivity) leaving 1-0.66=0.34 mol of NaBr

This will produce 0.66/2 = 0.33 mol of I2
and 0.34/2=0.17 mol Br2

Why have you done this? I don't understand.

Reply 6

oxymoron

7

Right, I'll try this from another angle.

In the solution you have 0.66 moles of NaBr and 0.66 moles of NaI

You are adding 0.5 moles of chlorine gas to this (which will ALL react)

For a moment, imagine you had ONLY 0.66 moles of NaI (no NaBr). In this case, 0.33 moles of chlorine will react with the 0.66 moles of NaI (because it is a 1:2 stoichiometry in the equation).

This leaves 0.17 moles of the original 0.5 moles of chlorine remaining (unreacted).

Now bring back the NaBr we forgot about and say the remaining 0.17 moles of chlorine will react with as much of the 0.66 moles NaBr as it can.
This happens to be 0.17 x 2 = 0.34 moles (again 1:2 relationship). So there is not enough chlorine to react with all the NaBr (there were 0.66 moles but only 0.34 moles can react).

The reason I said the chlorine reacts with the NaI first is because this reaction is preferable because there is a bigger reactivity gap between chlorine and Iodine than between chlorine and bromine .... SO
to summarise,

The chlorine reacts with as much of the NaI as possible, but when this supply is used up, any remaining chlorine is reacting with the NaBr.

To finish the question, we need to know the moles of gasses produced. (remember one mole of chlorine produces 1 mole of gas ... from the equation)
So we said 0.33 moles of chlorine reacted with NaI ... producing 0.33 moles of I2
and 0.17 moles reacted with NaBr ... producing 0.17 moles of Br2

I think that is right, but anybody feel free to correct me if it's not.
If you still don't understand ... which bit is unclear?

AS chem

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