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Kitty F.

asked • 11/07/14

How many grams of Lithium Nitrate are needed to make 250 grams of Lithium Sulfate?

Could someone please look over my work?  I had to research Stoichiometry as a self-study and I'm not sure if I'm on the right track.
The question is:
How many grams of Lithium Nitrate are needed to make 250 grams of Lithium Sulfate, assuming you have an adequate amount of Lead IV Sulfate to do the reaction? 
I believe Lead IV Sulfate is the limiting Reagent...As I understand it, the presence of this reagent is necessary for the reaction to occur and it is totally consumed when the reaction is complete.  An insufficient quantity of the limiting reagent would halt the reaction.
Pb(SO4)2 + 4LiNO3  -> Pb(NO3)4 + 2LiSO4
I've checked the equation to ensure it is balanced.  The reactants and products are present in equivalent amounts.  Reactants:  Pb (Lead): 1, S (Sulfur):  2, O (Oxygen):  20, Li (Lithium):  4, N (Nitrogen):  4; Products:  Pb (Lead): 1, S (Sulfur): 2, O (Oxygen): 20, Li (Lithium): 4, N (Nitrogen): 4
Pb(SO4)2  Lead IV Sulfate
Pb:  207.2
S:  32.07(2) -> 64.14
O  1600(4)  -> 64.00 (2)= 128
=399.34 g/mol
LiNO3  Lithium Nitrate
Li:  6.941
N:  14.01
O:  16.00(3) -> 48.00
=68.951 g/mol
Pb (NO3)4  Lead IV Phosphate
Pb:  207.2
N:  14.01(4)-> 56.04
O:  16.00(12) ->192
=455.24 g/mol
LiSO4 Lithium Sulfate
Li:  6.941
S:  32.07
O:  16.00 (4) -> 64.00
= 103.011 g/mol
Ratio of LiNO3 to LiSO4 (Lithium Nitrate:  Lithium Sulfate) 
4LiNO3: 2Li2SO4.  The ratio is 4:2, so we will need 4 moles of Lithium Nitrate to react completely with Lead IV Sulfate to produce 2 moles of Lithium Sulfate.
1 mole of Lithium Nitrate is 68.951 g/mol.
1 mole of Lithium Sulfate is 103.11 g/mol.
250 grams (given value from word problem) /68.951 g/mol= 3.62 moles
3.62 moles Lithium Nitrate/4 moles= .905 x 103.11 g/mol= 93.3 grams Lithium Nitrate
ANSWER:  93.3 grams Lithium Nitrate are needed.

Tom P.

Doesn't look correct--why would you divide the 250 grams of lithium sulfate you want to make by the molecular mass of lithium nitrate? Use the mass of 1 mole of lithium sulfate to get the number of moles you want to produce and then solve for moles and mass of lithium nitrate (by multiplying by the mass of lithium nitrate, or 68.951, not 103.11 as was done above).


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