WKS
Solution Stoichiometry 2
1.
Name Answer Key
Period
What volume of 0.325 M Na3PO4 would be needed to precipitate 25.00 g Ba3(PO4)2 with excess BaCl2
by the following balanced reaction?
2Na3PO4(aq) + 3BaCl2(aq) → 6NaCl(aq) + Ba3(PO4)2(s)
? mL Na 3PO 4 = 25.00 g Ba 3 (PO 4 )2 ×
1 mol Ba 3 (PO 4 )2
×
2 mol Na 3PO 4
×
1000 mL Na 3PO 4
601.9 g Ba 3 (PO 4 )2 1 mol Ba 3 (PO 4 )2 0.325 mol Na 3PO 4
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0.04154 mol
Ba 3 ( PO4 )2
0.08307 mol Na 3PO4
= 256 mL Na 3PO 4 (0.256 L)
2.
How many grams of CaO are required for complete reaction with the HCl in 275 mL of a 0.523 M HCl
solution? The balanced 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
3. When 53.0 mL of 0.750 M cobalt (III) nitrate are added to a sodium sulfate solution, how many grams
of cobalt (III) sulfate can be precipitated? Balance the equation first.
2 Co(NO3)3 (aq) + 3 Na2SO4 (aq) →
Co2(SO4)3 (s) + 6 NaNO3 (aq)
MM (Co2 (SO 4 )3 ) = 2 × 58.93 g + 3× 32.06 g + 12 × 16.00 g = 406.0 g/mol
0.750 mol Co(NO3 )3 1 mol Co2 (SO 4 )3 406.0 g Co2 (SO 4 )3
? g Co2 (SO 4 )3 = 53.0 mL Co(NO3 )3 ×
×
×
1000 mL Co(NO3 )3 2 mol Co(NO3 )3
1 mol Co2 (SO 4 )3
!############################" !###########"
0.0398 mol Co(NO3 )3
0.0199 mol Co2 (SO4 )3
= 8.07 g Co2 (SO 4 )3
4.
HC2H3O2 (acetic acid) is used to neutralize (deactivate) NaOH (sodium hydroxide) according to the
following balanced equation:
HC2H3O2(aq) + NaOH(aq) → NaC2H3O2(aq) + H2O(l)
a. How many moles of acetic acid are needed to neutralize 25.19 mL of 0.4295 M NaOH?
0.4295 mol NaOH 1 mol HC2 H 3O 2
? mol HC2 H 3O 2 = 25.19 mL NaOH ×
×
= 0.01082 mol HC2 H 3O 2
1000 mL NaOH
1 mol NaOH
!#############"
0.01082 mol NaOH
b. If the volume of the acetic acid solution used is 34.57 mL, what is it concentration, in M?
0.01082 mol HC2 H 3O 2
1 mL
M=
×
= 0.3130 M HC2 H 3O 2
34.57 mL
1× 10−3 L
5.
HCl (hydrochloric acid) reacts with Na2CO3 (sodium carbonate) to form NaCl (sodium chloride),
water, and CO2 (carbon dioxide
2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + H2O(l) + CO2(g).
a. How many moles of HCl are required to form 11.1 grams of CO2?
? mol HCl = 11.1 g CO 2 ×
1 mol CO 2
×
2 mol HCl
44.01 g CO 2 1 mol CO 2
!#########"
= 0.504 mol HCl
0.252 mol CO2
b. If 321 mL of HCl solution reacted, what was 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
6. Gravimetric analysis is a method of determining the concentration of a compound in solution by
measuring the mass of a precipitate. In one experiment, 1.18 g AgCl precipitates when 25.0 mL of
AgNO3 solution reacts with excess NaCl solution in the following reaction:
1 AgNO3 (aq) + 1 NaCl (aq) → 1 NaNO3 (aq) + 1 AgCl (s)
a. Balance the equation.
b. How many moles of AgNO3 were reacted?
MM (AgCl) = 107.87 g + 35.45 g = 143.3 g/mol
1 mol AgCl 1 mol AgNO3
? mol AgNO3 = 1.18 g AgCl ×
×
= 8.23× 10−3 mol AgNO3
143.3 g AgCl 1 mol AgCl
!##################
"
−3
8.23×10
mol AgCl
c. What is the concentration of AgNO3 solution, in M?
8.23× 10−3 mol AgNO3
1 mL
M (AgNO3 ) =
×
= 0.329 M
25.0 mL
1× 10−3 L