Question

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Stoichiometry LR and XSR
(a) If you have in mind the reaction below with its reactants to produce 12.0g of lead (IV) chloride, can you do this with 10.45g of lead (IV) sulfate solution and 13.86g of lithium chloride? Either way (yes, you can or no, you cannot)
(b) identify if there is a limiting reactant and an excess reactant - which reactant is which?
(c) determine (calculate) if there is any mass of the reactants leftover that did not react with other reactant

Answer 1

a) No \[4LiCl + Pb(SO4) \rightarrow 2Li2SO4 + PbCl4\]
4:1 reaction ratio between Lithium Chloride and Lead(IV) Sulfate, so

b) Lithium Chloride is the limiting reactant

c) 13.86g LiCl / 4 (4:1 ratio) = 3.465g of Pb(SO4)2 reacted, 6.985g remaining

hope this is right, it's been a while

Answer 2

I love you man @tangowilde

Answer 3

Part of this is incorrect. When doing calculations like this you need to work in moles. (because mass is not equivalent to the number or particles/molecules between substances).

The first answer is correct.

b) is incorrect.

\(n_{LiCl}=\dfrac{13.86~g}{42.394~g/mol}=0.32693~ mol\)

\(n_{Pb(SO_4)_2}=\dfrac{10.45~g}{271.332~g/mol}=0.03851 ~mol\)

so in fact the limiting reactant is \(Pb(SO_4)_2\).

c) is also incorrect, you need to work in moles.

Answer 4

aw heck, i'm sorry. it's been a long time. so 0.1729 moles or 7.32g of LiCl remain?
I used 399.3252 g/mol for Pb(SO4)2, i wasn't sure how you got 271. correct me if wrong please!

Answer 5

oh damn. yep, i used the wrong molar mass, I'm too lazy to calculate molar masses so i typed into google, didn't notice it gave me lead sulfide.

I just noticed i made a mistake too. I forgot to divide the moles by it's stoichiometric coefficients to find the limiting reactant. But the answer is still \(Pb(SO_4)_2\).

Answer 6

I kinda turned this in for formative points..I wish you could've commented yesterday..