Excess and Limited...

[jack_chay]

when given the formula:
5Cr203 + 6KBr03 +4KOH ---> 5K2Cr2O7 +2H20 +3Br2

how do you work out step by step which is the limiting agent?

Thanks in advance

[Stevensmay]

Unless I am missing something, or this is in a different context, you also need to know how much mole of each reactant you have. Without it it is impossible to say which reactant is limited and which is in excess.

If these are not provided then I am wrong.

[jack_chay]

opps, sorry I forgot to add mol of Cr2O3 = 0.1 and mol of KBrO3 = 0.09288

Thank you

[RKTR]

opps, sorry I forgot to add mol of Cr2O3 = 0.1 and mol of KBrO3 = 0.09288

Thank you

from the question , we can see that we need 6 mol of KBrO3 for 5 mol of Cr2O3
we have 0.1mol of Cr2O3 , we need 0.12 mol of KBrO3 to make sure everything completely react. therefore, the KBrO3 is the limiting reagent since there' s only 0.09288

[Stevensmay]

So for this type of question, it is done in two parts.

n(Cr2O3) = 0.1
To find n(KbrO3) we simply do n(Cr2O3) * want\got
n(KbrO3) = .1 * 6/5 = .12

Since this is higher than the actual n(KbrO3) we have, which is 0.09288, the Cr2O3 must be in excess as it will not run out.

If we did this for the KBrO3 we would find that only 0.0774 mole of Cr2O3 would be reacted with, as we would run out of KBrO3.

From this we know that KBrO3 is the limiting reactant and Cr2O3 is in excess.

[jack_chay]

so would you say that it is in excess by 0.0226 mol?

[Stevensmay]

That Cr2O3 is in excess by 0.0226 mol, so yes.

[jack_chay]

im sorry it seems like I ask a lot, but in

N2O3 +2NaOH ---> ..........................

N2O3 has a mol of 0.190789474
NaOH has a mol of 0.08085

Is it N2O3 in excess by 0.300728948?

Thank you

Thank you

[Stevensmay]

Yes

[jack_chay]

yup thank you for helping me in chemistry too

[jack_chay]

I was told not to start another topic, so is this how i'm meant to ask another question?

23.9 mL of 0.704 M NaOH solution is added
to 37.7 mL of 1.274 M KOH solution and the
resulting mixture is just neutralised by 30.8 mL
of H2SO4 solution

Write an ionic equation for the neutralisation
reaction.

Thanks in advance

[Stevensmay]

I think you want

H+ + HSO4- + 2K+ + 2OH- -> 2K+ + SO4^2- + 2H2O ionic equation
To do this you essentially split each compound up into it's molecular components, giving them charges.
Would type this out but is going to get very messy.
Does anyone know who to speak to regarding another latex package installed?

NaOH + KOH + H2SO4 -> Na2SO4 + H20 + K molecular equation
Make sure to include states!

[jack_chay]

oh ok I see... thank you a million times