NAME _________________________________ Chem. 163 Class Problems #1 Score ____/40
These are review problems from past concepts that you need to recall for success in this term of chem 163
I. GAS PROBLEM (5 pts) A sample of solid potassium chlorate is heated in a test tube and decomposed.
The oxygen produced in the reaction is collected over water at 22oC and the Ptot is 754 torr. The gas volume
collected totals 0.650 L. Calculate the partial pressure of oxygen gas collected and the mass of sample that
was decomposed.
2 KClO3 (s) ---- 2 KCl (s) + 3 O2 (s)
Ptot = 754 torr
Vapor pressure H2O @ 22oC = 19.8 torr
PO2 = 754 – 19.8 = 734.2 torr [734.2/760 = 0.966 atm]
Partial Pressure O2= 734.2 torr (0.966 atm)
V = 0.650 L
PV = nRT
P = 0.966 atm
T = 22 + 273 = 295 K
nKClO3 = 122.6 g/mol
moles O2 = (PV)/(RT) = [(0.966 atm)*(0.650 L)] / [(0.082056 Latm/molK)*(295 K)]
= [0.6279] / [24.20652] = 0.0259 mols O2
Find mass reactant, KClO3 convert moles O2 to moles KClO3 then use formula wt convert to mass, g
(0.0259 mols O2)* (2 mols KClO3/3 mols O2) = 0.0173 mols KClO3
(0.0173 mols)*(122.6 g/mol) = 2.12 gKClO3
II. REACTION TYPES. (30 pts) Identify the type of reaction, complete, balance, and other questions as
asked. If there is a NR, state the reason why.
1) C
2 -3 - 2 Al (s) + 3 Ba(ClO3)2 (aq)
TOT: 2 Al+3(s) + 6 ClO3-1(aq) + 3 Ba0(s) 2 Al0(s) + 3 Ba+2(s) + 6 ClO3-1(aq)
NET: 2 Al+3(s) + 3 Ba0(s) 2 Al0(s) + 3 Ba+2(s)
2) D
reduced: Al
2- 3 --- Fe2S3(s) + 6 NaNO3(aq)
3) C 1 – 1 --- ZnSO4(aq) + H2(g)
TOT: 1 Zn0(s) + 2 H+1(aq) + 1 SO4-2(aq) 1 Zn+2(aq) + 1 SO4-2(aq) + 1 H20(g)
NET: Zn0(s) + 2 H+1(aq) Zn+2(s) + H20(g)
ox ag: H
red ag: Zn
oxidized: Ba
4) E, no rxn Ba not active enough to replace K in a cmpd
5) B
1 -- 1 SrCl2(aq) + 3 O2(aq)
6) Refer to Part II, #3. On the following conditions: 35.6 g zinc reacts with 2.59 mols of acid.
a) By calculations, find the amount of mass of product formed from each reactant.
1 Zn (s) + 1 H2SO4 (aq)  1 ZnSO4 (aq) + 1 H2(g)
Zinc:
Zn = 65.4 g/mol
(35.6 g)*(65.4 g/mol) = 0.544 mol Zn
mass ZnSO4 [161.5 g/mol]: (0.544 mol Zn)*(1 mol ZnSO4/1 mol Zn) = 0.544 mol ZnSO4
(0.544 mol)*(161.5 g/mol) = 87.9 g ZnSO4
mass H2 [2.0 g/mol]: (0.544 mol Zn)*(1 mol H2/1 mol Zn) = 0.544 mol H2
(0.544 mol)*(2.0 g/mol) = 1.09 g H2
Acid:
2.59 mols
H2SO4 = 98.1 g/mol
mass ZnSO4 [161.5 g/mol]: (2.59 mol H2SO4)*(1 mol ZnSO4/1 mol H2SO4) = 2.59 mol ZnSO4
(2.59 mol)*(161.5 g/mol) = 418.3 g ZnSO4
mass H2 [2.0 g/mol]: (2.59 mol H2SO4)*(1 mol H2/1 mol H2SO4) = 2.59 mol H2
(2.59 mol)*(2.0 g/mol) = 5.18 g H2
b) Determine and identify the Limiting Reagent
Zn produces least amount of product
c) Once you have found the Limiting Reagent, how much of the other reactant is still left.
What is the most, maximum, amount of product that can be produced?
Max pdt formed: ZnSO4 = 87.9 g
H2 = 1.09 g
Amt H2SO4 left = Initial Amt – Consumer Amt
Initial Amt: (2.59 mol)*(98.1 g/mol) = 254.1 g
Consumed Amt: (0.544 mol Zn)*(1 mol H2SO4/1 mol Zn) = 0.544 mol H2SO4
(0.544 mol)*(98.1 g/mol) = 53.4 g H2SO4
254.1 – 53.4 = 200.7 g H2SO4 left
III. OXIDATION NUMBERS (5 pts) Calculate the “oxidation” number for each element in the compound.
1) Al2(SiO3)3
+3 +4 -2
2) OF2
+2 -1