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
S: 32.07(2) -> 64.14
O 1600(4) -> 64.00 (2)= 128
LiNO3 Lithium Nitrate
O: 16.00(3) -> 48.00
Pb (NO3)4 Lead IV Phosphate
N: 14.01(4)-> 56.04
O: 16.00(12) ->192
LiSO4 Lithium Sulfate
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.