Limiting reagent (1 Viewer)

[highshill]

https://thsconline.github.io/s/?vie...KQkVTTFR6S0pKa3c&n=Baulkham Hills 2012 w. sol
I am a bit confused on 23 (c) Where did I go wrong:

My solution:
n=m/M finding moles of Pb(NO3)2
20.0g/331.22
=0.06038282712g/mol

Finding KI moles

n=m/M
20/166
O.1204819277


So isn't Pb(NO3)2 the limiting reagent. why does the answer say that KI is the limiting reactant where did I go wrong?

 

[dan964]

https://thsconline.github.io/s/?vie...KQkVTTFR6S0pKa3c&n=Baulkham Hills 2012 w. sol
I am a bit confused on 23 (c) Where did I go wrong:

My solution:
n=m/M finding moles of Pb(NO3)2
20.0g/331.22
=0.06038282712g/mol

Finding KI moles

n=m/M
20/166
O.1204819277


So isn't Pb(NO3)2 the limiting reagent. why does the answer say that KI is the limiting reactant where did I go wrong?

I think that is a mistake in 23(a). The equation is not balanced. Should be 2KI not KI.

I have updated the file above, with the correction.


Also added this note:

Note on Q23(b) The explanation isn't super clear.

While the number of moles is less for Pb(NO3)2.

Twice as many moles of KI are used in the process.

If I had 0.12048 moles of KI, this would mean I need 0.06024 moles which is less than I have
so I have more Pb(NO3)2 than I need.

On the other hand, if I had 0.06038 moles of Pb(NO3)2, I need 0.12076 moles
which is more than I have. Hence KI is a limiting reagant


~dan

 

[He-Mann]

The stoichiometric ratio should be 1:2:1:2... This school has lost all credibility.

This means you would need to halve the n(KI) (or double n(Pb(NO3)2) if you want to compare like that) (WHY?) which leads to KI being the limiting reagent.

Essentially, the balance equation tells us that 1 mole of Pb(NO3)2 is required to react with 2 moles of KI for the reaction to proceed. In other words, the mole ratio between KI and Pb(NO3)2 is 2:1.

Using your calculations, n(KI)/n(Pb(NO3)2) = 1.99.. < 2. This implies KI is the limiting reagent. (WHY?)


"you need to act know to improve your capabilities here" - how can you make this mistake and teach at this level, at this school...

 

[highshill]

i agree

 

[He-Mann]

i agree

Have you resolved your problem?