AP CHEMISTRY WKST: MOLAR STOICHIOMETRY
p. 1
2Na3PO4(aq) + 3Ba(NO3)2(aq)  Ba3(PO4)2(s) + 6NaNO3(aq)
1.
If 350.0 mL of a 0.85 M Na3PO4 solution reacts, what mass of barium phosphate forms?
 0.85 mol Na 3PO4
350 .0 mL Na 3PO4 
 1000 mL Na 3PO4
2.
 1 mol Ba3 (PO4 )2

 2 mol Na PO
3
4

 3 mol Ba(NO 3 )2

 1 mol Ba (PO )
3
4 2

 1000 mL Ba(NO 3 )2

 1.500 mol Ba(NO )
3 2


  166 mL Ba(NO 3 )2 solution


What molarity of sodium phosphate is needed in order to have 100.0 mL of it react with 150.0 mL of a 1.25 M barium
nitrate solution?
 1.25 mol Ba(NO 3 )2
150 .0 mL Ba(NO 3 )2 
 1000 mL Ba(NO 3 )2
M
4.

  90. g Ba3 (PO4 )2


How many mL of a 1.500 M Ba(NO3)2 solution is needed to react to form 50.0 g of barium phosphate?
 1 mol Ba3 (PO4 )2
50.0 g Ba3 (PO4 )2 
 601 .93 g Ba3 (PO4 )2
3.
 601 .93 g Ba3 (PO4 )2

 1 mol Ba (PO )
3
4 2

 2 mol Na 3PO4

 3 mol Ba(NO )
3 2


  0.125 mol Na 3PO4


mol solute
0.125 mol Na 3PO4

 1.25 M Na 3PO4
L so ln
0.1000 L
If 500.0 mL of a 0.750 M Na3PO4 solution is mixed with 250.0 mL of a 1.75 M Ba(NO3)2 solution:
a) what mass of Ba3(PO4)2 will form?
 0.750 mol Na 3PO4
500 .0 mL Na 3PO4 
 1000 mL Na 3PO4
 1 mol Ba3(PO4 )2  601 .93 g Ba3(PO4 )2 



 2 mol Na PO  1 mol Ba (PO )   113 g Ba3(PO4 )2
3
4 
3
4 2


 1.75 mol Ba(NO 3 )2
250 .0 mL Ba(NO 3 )2 
 1000 mL Ba(NO 3 )2
ANSWER: 87.8 g Ba3(PO4)2
 1 mol Ba3 (PO4 )2

 3 mol Ba(NO )
3 2

 601 .93 g Ba3 (PO4 )2

 1 mol Ba (PO )
3
4 2


  87.8 g Ba3 (PO4 )2


AP CHEMISTRY WKST: MOLAR STOICHIOMETRY
p. 2
b) what is the concentration of the remaining ions in solution?
Note: From #4a, Ba(NO3)2 is the limiting reactant.
M1V1 = M2V2
[
[
]
(
)]
amount of PO43− reacted:
(
+
Na
PO43−
Ba2+
−
NO3
)
[ ]i
1.50
0.500
0.583
1.17
Δ[]
0
−0.389
−0.583
0
[ ]f
1.50 M
0.111 M
--1.17 M
This next part is an alternate method of doing #4b. You find the moles of each reactant using mol=M∙
MOLES METHOD:
mol Na3PO4 = (0.750 M)(0.5000 L) = 0.375 mol Na 3PO4
mol Ba(NO3)2 = (1.75 M)(0.2500 L) = 0.438 mol Ba(NO3)2
amount of PO43− reacted:
 2 mol PO34 
0.438 mol Ba2  
 3 mol Ba2 


  0.292 mol PO34 


IN SOLUTION
Na+
PO43−
Ba2+
−
NO3
moli
1.125
0.375
0.438
0.876
Δ mol
0
−
−
0
molf
1.125
0.083
0
0.876
[ ]
1.50 M Na+
0.11 M PO43−
−−−
−
1.17 M NO3
NOTE: to find the M of the ions, divide the molf by 0.7500 L (which is the combined volume of the two solutions)
AP CHEMISTRY WKST: MOLAR STOICHIOMETRY
5.
What volume of a 0.250 M HCl solution is required to completely react with the following bases?
MH  VA  M OH VB
VA 
M OH VB
MH
a) 60.0 mL of a 0.250 M NaOH solution
VA 
(0.250 M)(60.0 mL)
 60.0 mL HCl solution
0.250 M
b) 120.0 mL of a 0.150 M KOH solution
VA 
(0.150 M)(120 .0 mL)
 72.0 mL HCl solution
0.250 M
c) 50.00 mL of a 0.250 M Ca(OH)2 solution
NOTE: [OH−] = 0.500 M
VA 
(0.500 M)(50.0 mL)
 100 .0 mL HCl solution
0.250 M
p. 3