WKS – Honors
Solution Stoichiometry
Name Answer Key
Period
1) How many mL of 0.348 M Hg(NO3)2 are needed to fully react with 15.0 mL of 0.485 M KBr according
to the following unbalanced reaction?
Hg(NO3)2(aq) + 2 KBr(aq) →
HgBr2(s) + 2 KNO3(aq)
0.485 mol KBr 1 mol Hg(NO3 )2 1000 mL Hg(NO3 )2
? mL Hg(NO3 )2 = 15.0 mL KBr ×
×
×
1000 mL KBr
2 mol KBr
0.348 mol Hg(NO3 )2
!##########" !############"
0.007275 mol KBr
0.03638 mol mol Hg(NO3 )2
= 10.5 mL Hg(NO3 )2
2) What volume of 0.325 M Na3PO4 would be needed to fully react with 25.0 mL of 0.480 M BaCl2 by the
following unbalanced reaction?
2 Na3PO4(aq) + 3 BaCl2(aq) → 6 NaCl(aq) +
Ba3(PO4)2(s)
? mL Na 3PO4 = 25.0 mL BaCl 2 ×
0.480 mol BaCl 2
×
2 mol Na 3PO4
1000 mL BaCl 2
3 mol BaCl 2
!############" !##########"
0.0120 mol BaCl 2
×
1000 mL Na 3PO4
0.325 mol Na 3PO4
0.00800 mol Na 3PO4
= 24.6 mL Na 3PO4
3) Calculate the volume of 0.2250 M HC2H3O2 (acetic acid) solution needed to neutralize 25.19 mL of
0.4295 M NaOH in the following unbalanced reaction.
HC2H3O2(aq) +
NaOH(aq) →
NaC2H3O2(aq) +
H2O(l)
0.4295 mol NaOH 1 mol HC 2 H 3O2
1000 mL HC 2 H 3O2
? mL HC 2 H 3O2 = 25.19 mL NaOH ×
×
×
1000 mL NaOH
1 mol NaOH
0.2250 mol HC 2 H 3O2
!#############" !############"
0.01082 mol NaOH
0.01082 mol HC 2 H 3O2
= 48.08 mL HC 2 H 3O2
4) What volume of 0.325 M Na3PO4 would be needed to precipitate 25.00 g Ba3(PO4)2 with excess BaCl2
by the following unbalanced reaction?
2 Na3PO4(aq) + 3 BaCl2(aq) → 6 NaCl(aq) +
Ba3(PO4)2(s)
? mL Na 3PO4 = 25.00 g Ba 3 (PO4 )2 ×
1 mol Ba 3 (PO4 )2
×
2 mol Na 3PO4
601.9 g Ba 3 (PO4 )2 1 mol Ba 3 (PO4 )2
!#############" !############"
0.04154 mol
Ba 3 ( PO4 )2
= 256 mL Na 3PO4 (0.256 L)
0.08307 mol Na 3PO4
×
1000 mL Na 3PO4
0.325 mol Na 3PO4
5) How many grams of CaO are required for complete reaction with the HCl in 275 mL of a 0.523 M HCl
solution? The unbalanced equation for the reaction is:
CaO(s) + 2 HCl(aq) →
CaCl2(aq) +
H2O(l)
0.523 mol HCl 1 mol CaO 56.1 g CaO
? g CaO = 275 mL HCl ×
×
×
= 4.03 g CaO
1000 mL HCl 2 mol HCl 1 mol CaO
!##########" !########"
0.144 mol HCl
0.0719 mol CaO
6) 34.57 mL of HC2H3O2 solution of unknown concentration is used to neutralize 25.19 mL of NaOH with
concentration 0.4295 M according to the following unbalanced equation:
HC2H3O2(aq) +
NaOH(aq) →
NaC2H3O2(aq) +
H2O(l)
a) How many moles of acetic acid are used?
0.4295 mol NaOH 1 mol HC 2 H 3O2
? mol HC 2 H 3O2 = 25.19 mL NaOH ×
×
= 0.01082 mol HC 2 H 3O2
1000 mL NaOH
1 mol NaOH
!#############"
0.01082 mol NaOH
b) What is the concentration of the acetic acid solution, in M?
0.01082 mol HC 2 H 3O2
1 mL
M=
×
= 0.3130 M HC 2 H 3O2
34.57 mL
1 × 10 −3 L
7) When 321 mL of HCl solution of unknown concentration reacts with Na2CO3, it forms NaCl, water, and
11.1 g of CO2:
2 HCl(aq) +
Na2CO3(aq) → 2 NaCl(aq) +
H2O(l) +
CO2(g).
a) How many moles of HCl are used in the reaction?
1 mol CO2
2 mol HCl
? mol HCl = 11.1 g CO2 ×
×
= 0.504 mol HCl
44.01 g CO2 1 mol CO2
!#########"
0.252 mol CO2
b) What is the concentration of the HCl solution, in M?
0.504 mol HCl
1 mL
M=
×
= 1.57 M HCl
321 mL
1 × 10 −3 L