REVIEW PROBLEMS
CHM131
SINGLE STEP PROBLEMS
These problems are based on simple single step conversions.
1. What is the density of a material that has a mass of 34.5 g and a volume of 1.45 ml?
m
34.5 g
d = ----- = --------------- = 2.38 g/ml
V
1.45 ml
2. What is the mass of 3.4 ml of a material with a density of 6.89 g/ml?
m = d × V = (6.89 g/ml) × 3.4 ml = 23 g
3. What is the volume of a 100.4 g of a fluid that has a density of 6.65 g/ml?
V = m/d = (100.4 g)/(6.65 g/ml) = 15.1 ml
4. How many moles are there in 34.85 g of Ammonium Chloride?
?mol NH4Cl = 34.85 g NH4Cl × (1 mol NH4Cl/53.492 g NH4Cl) = 0.6515 mol NH4Cl
5. How many grams are needed for 2.90 moles of Barium Sulfate?
?g BaSO4 = 2.90 mol BaSO4 × (233.37 g BaSO4/1 mol BaSO4) = 677 g BaSO4
6. How many grams of Barium Nitrate are present in a 12.5% 500 g solution?
Percent Composition = 100% × (mass solute/mass solvent)
Mass solvent × (percent comp./100%) = mass solute = 500 g × (12.5%/100%) = 62.5 g Ba(NO3)2
7. How many moles of Sodium Sulfate are present in 0.35 l of a 0.75 M solution?
M = mol/V => mol = M × V =(0.75 M) × (0.35 l) = 0.26 moles Na2SO4
8. What molarity is a solution made of 4.5 moles of sodium sulfate dissolved in a solution of 0.200 l?
M = mol/V = (4.5 moles)/(0.200 l) = 0.9 M Na2SO4
9. What is the volume of a 2.4 M solution that is made with 8.0 moles of sodium sulfate?
M = mol/V => V = mol/M = 8.0 mol/2.4 M = 3.3 l
10. What is the molarity of a 2.5 M solution that is diluted to 2.4 times its original volume?
M1V1 = M2V2
2.5 M (1) = M2(2.4)
M2 = (2.5 M × 1)/2.4 = 1.0 M
11. What is the volume of solvent needed to dilute a 100 ml 3.4 M solution to 0.75 M?
M1V1 = M2V2
3.4 M × 100 ml = 0.75 M × V2
V2 = (3.4 M × 100 ml)/0.75 M = 453 ml
TWO STEP PROBLEMS
It is time to take it up a notch and try some problems that require more than a single step.
1. What is the molarity of a solution containing 2.5 moles of sodium sulfate in a total volume of 200 ml?
Convert 200 ml to liters = 0.200 l (if fuzzy on this review chapter 1)
M = mol/liter = 2.5 mol Na2SO4/0.200 l = 12.5 M Na2SO4
2. How many grams of Ammonium Chloride are there is 1.2 l of a 2 M solution?
?g NH4Cl = (2 mol/l) × (1.2 l) × (53.492 g NH4Cl/1 mol) = 100 g NH4Cl
3. If you have 3.4 l of Oxygen gas, at a density of 23.0 g/l, how many moles of Oxygen gas do you have?
? mol O2 = 3.4 l O2 × (23.0 g/l) × (1 mol O2/32.00 g O2) = 2.4 mol O2
4. How many moles of Barium Nitrate are present in a 12.5% 500 g solution?
Percent Composition = 100% × (mass solute/mass solvent)
Mass solvent × (percent comp./100%) = mass solute = 500 g × (12.5%/100%) = 62.5 g Ba(NO3)2
?mol Ba(NO3)2 = 62.5 g Ba(NO3)2 × (1 mol Ba(NO3)2/261.32 g Ba(NO3)2) = 0.239 mol Ba(NO3)2
BALANCING EQUATIONS
The last fifteen questions have been pretty straight forward, now we need to add some equation
balancing in, since that is the core of any stoichiometry. Balance the following skeleton equations.
1. Ag(s) + Cl2(g)  AgCl(s)
2 Ag(s) + Cl2(g) 2 AgCl(s)
2. AgNO3(aq) + BaCl2(aq)  AgCl(s) + Ba(NO3)2(aq)
2 AgNO3(aq) + BaCl2(aq)  2AgCl(s) + Ba(NO3)2(aq)
3. C2H5OH(l) + O2(g)  CO2(g) + H2O(g)
2 C2H5OH(l) + 6 O2(g) 4 CO2(g) + 6 H2O(g)
4. K3N(aq) + Hg(NO3)2(aq)  KNO3(aq) + Hg3N2(s)
2 K3N(aq) + 3 Hg(NO3)2(aq)  6 KNO3(aq) + Hg3N2(s)
PREDICTING REACTIONS
It is one thing to balance an equation. It is another to predict a reaction based on the known rules of
reactivity. Given the following scenarios, predict the balanced equation for the reaction.
1. Mixing aqueous solutions of Lead(II) Nitrate and Potassium Chloride.
Pb(NO3)2(aq) +2 KCl(aq)  PbCl2(s) +2 KNO3(aq) (precipitate reactions)
2. Combusting Silver metal in the presence of Oxygen gas.
4 Ag(s) + O2(g)  2 Ag2O(s)
3. The decomposition of Ammonium Sulfate in the presence of heat.
(NH4)2SO4(s)  2 NH3(g) + H2SO4(l) (decomposition of Ammonium compounds)
4. The decomposition of Barium Carbonate in the presence of heat.
BaCO3(s)  BaO(s) + CO2(g) (decomposition of carbonates)
5. The addition of Barium Sulfide to a solution of hydrochloric acid.
BaS(s) + 2 HCl(aq)  BaCl2(aq) + H2S(g) (gas formation rules)
6. The mixture of Barium Hydroxide and Perchloric acid solutions.
Ba(OH)2(aq) + 2 HClO4(aq)  Ba(ClO4)2(aq) + 2 H2O(l) (acid/base neutralization)
LIMITING REAGENTS
Now it is time to put this together a little. You need to determine the limiting reagent in each of the
following equations.
1. 46.9 g Sulfur Dioxide mixed with 35.0 g Oxygen Gas.
Reaction: 2 SO2(g) + O2(g)  2 SO3(g)
?moles SO2 = 46.9 g × (1 mol/64.07 g) = 0.732 mol SO2
?moles O2 = 35.0 g × (1 mol/32.00 g) = 1.09 mol O2
Since SO2:O2 is 2:1, SO2 must be limiting.
2. 12.3 g Hydrogen Gas combusted with 78.4 g Oxygen Gas.
Reaction: 2 H2(g) + O2(g)  2 H2O(l)
?moles H2 = 12.3 g × (1 mol/2.016 g) = 6.10 mol O2
?moles O2 = 78.4 g × (1 mol/32.00 g) = 2.45 mol O2
Ratio in equation for H2:O2 = 2:1, ration here is ≈ 3:1, so Oxygen must be limiting.
3. 500 ml of a 2M HCl solution combined with 750 ml of a 1.8M NaOH solution
Reaction: HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
? moles HCl = (2 mole/l )× 0.500 l = 1 mol HCl
? moles NaOH = (1.8 mole/l )× 0.750 l = 1.35 mol NaOH
Ratio in equation for HCl:NaOH = 1:1, ration here is ≈ 1:1.35, so HCl must be limiting.
4. 200.0 g of silver metal decomposed with 45.0 g Oxygen gas.
Reaction: 4 Ag(s) + O2(g)  2 Ag2O(s)
?mol Ag = 200.0 g × (1 mol/107.9 g) = 1.854 mol Ag
?moles O2 = 45.0 g × (1 mol/32.00 g) = 1.41 mol O2
Ratio in equation for Ag:O2 = 4:1, ration here is ≈ 1.3:1, so Ag must be limiting.
5. A 200.0 ml volume of sulfur dioxide at 2.4 g/ml and 100 ml of Oxygen Gas at a density of 4.6 g/ml.
Reaction: 2 SO2(g) + O2(g)  2 SO3(g)
?mol SO2 = 200.0 ml × (2.4 g/ml) × (1 mol/64.07 g) = 7.5 mol SO2
?moles O2 = 100 ml × (4.6 g/ml) × (1 mol/32.00 g) = 14 mol O2
Since SO2:O2 is 2:1, SO2 must be limiting.