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Stoichiometry Practice Stoichiometry Practice ProblemProblem
Nick MaskreyNick Maskrey
ProblemProblem
If 5.26 g of Lithium Sulfate react with If 5.26 g of Lithium Sulfate react with Sodium Phosphate complete the Sodium Phosphate complete the stoichiometry.stoichiometry.
Complete and Balance the Complete and Balance the EquationEquation
3Li3Li22(SO4) + 2Na(SO4) + 2Na33(PO(PO44) ---> 2Li) ---> 2Li33(PO(PO44) + ) + 3Na3Na22(SO(SO44))
Split Each Chemical Into Separate Split Each Chemical Into Separate ColumnsColumns
3Li3Li22(SO4) + 2Na(SO4) + 2Na33(PO(PO44) ---> 2Li) ---> 2Li33(PO(PO44) + ) + 3Na3Na22(SO(SO44))
Write the given measurement into Write the given measurement into the correct columnthe correct column
3Li3Li22(SO4) + 2Na(SO4) + 2Na33(PO(PO44) ---> 2Li) ---> 2Li33(PO(PO44) + ) + 3Na3Na22(SO(SO44))5.26 g.
Convert the given into molesConvert the given into moles
3Li3Li22(SO4) + 2Na(SO4) + 2Na33(PO(PO44) ---> 2Li) ---> 2Li33(PO(PO44) + ) + 3Na3Na22(SO(SO44))5.26 g.
5.26g x 1 mole
1 109.945 g
2(6.941)
+ 32.065
+ 4(15.9994)
109.945 g
.0478 moles
Get moles for all other columns by putting the Get moles for all other columns by putting the moles of the given times (x) a fraction in eachmoles of the given times (x) a fraction in each
3Li3Li22(SO4) + 2Na(SO4) + 2Na33(PO(PO44) ---> 2Li) ---> 2Li33(PO(PO44) + ) + 3Na3Na22(SO(SO44))5.26 g.
5.26g x 1 mole
1 109.945 g
2(6.941)
+ 32.065
+ 4(15.9994)
109.945 g
.0478 moles
.0478 x / .0478 x / .0478 x /
The numerator is the coefficient of The numerator is the coefficient of the columnthe column
3Li3Li22(SO4) + 2Na(SO4) + 2Na33(PO(PO44) ---> 2Li) ---> 2Li33(PO(PO44) + ) + 3Na3Na22(SO(SO44))5.26 g.
5.26g x 1 mole
1 109.945 g
2(6.941)
+ 32.065
+ 4(15.9994)
109.945 g
.0478 moles
.0478 x 2/ .0478 x 2/ .0478 x 3/
The denominator is the coefficient The denominator is the coefficient of the givenof the given
3Li3Li22(SO4) + 2Na(SO4) + 2Na33(PO(PO44) ---> 2Li) ---> 2Li33(PO(PO44) + ) + 3Na3Na22(SO(SO44))5.26 g.
5.26g x 1 mole
1 109.945 g
2(6.941)
+ 32.065
+ 4(15.9994)
109.945 g
.0478 moles
.0478 x 2/3 .0478 x 2/3 .0478 x 3/3
Do all of the math to get the molesDo all of the math to get the moles
3Li3Li22(SO4) + 2Na(SO4) + 2Na33(PO(PO44) ---> 2Li) ---> 2Li33(PO(PO44) + ) + 3Na3Na22(SO(SO44))5.26 g.
5.26g x 1 mole
1 109.945 g
2(6.941)
+ 32.065
+ 4(15.9994)
109.945 g
.0478 moles
.0478 x 2/3
.0319 moles
.0478 x 2/3
.0319 moles
.0478 x 3/3
.0478 moles
Convert All to GramsConvert All to Grams3Li3Li22(SO4) + 2Na(SO4) + 2Na33(PO(PO44) ---> 2Li) ---> 2Li33(PO(PO44) + ) +
3Na3Na22(SO(SO44))5.26 g.
5.26g x 1 mole
1 109.945 g
2(6.941)
+ 32.065
+ 4(15.9994)
109.945 g
.0478 moles
.0478 x 2/3
.0319 moles
.0319 m x 163.939g
1 1 mole
3(22.989)
+ 30.973
+ 4(15.994)
163.938 g
5.23 g
.0478 x 2/3
.0319 moles
.0319 m x 115.994g
1 1 mole
3(6.941)
+ 30.973
+ 4(15.9994)
115.794 g
3.69 g
.0478 x 3/3
.0478 moles
.0478m x 142.041g
1 1 mole
2(22.989)
+ 32.065
+4(15.9994)
142.041 g
6.79 g
Verify the law of conservation of Verify the law of conservation of massmass
3Li3Li22(SO4) + 2Na(SO4) + 2Na33(PO(PO44) ---> 2Li) ---> 2Li33(PO(PO44) + ) + 3Na3Na22(SO(SO44))5.26 g.
5.26g x 1 mole
1 109.945 g
2(6.941)
+ 32.065
+ 4(15.9994)
109.945 g
.0478 moles
5.26g + 5.23g
10.49g
.0478 x 2/3
.0319 moles
.0319 m x 163.939g
1 1 mole
3(22.989)
+ 30.973
+ 4(15.994)
163.938 g
5.23 g
.0478 x 2/3
.0319 moles
.0319 m x 115.994g
1 1 mole
3(6.941)
+ 30.973
+ 4(15.9994)
115.794 g
3.69 g
3.69g + 6.79 g
10.48g
.0478 x 3/3
.0478 moles
.0478m x 142.041g
1 1 mole
2(22.989)
+ 32.065
+4(15.9994)
142.041 g
6.79 g
AnswerAnswer
The reactant mass adds up to 10.49 The reactant mass adds up to 10.49 g and the products mass adds up to g and the products mass adds up to 10.48 g. The third significant digit in 10.48 g. The third significant digit in each is 4 and within two of each each is 4 and within two of each other so the problem is correct.other so the problem is correct.
