Module Five Assignment
Chemical Equations
Instructions: Solve the following chemical equations.
1. For the following reaction, calculate how many moles of NO 2 forms when 0.356 moles of the
reactant completely reacts.
2 N 2 O5( g) → 4 NO2( g )+O2( g)
2 moles of N2O5 for every 4 moles of NO2
0.356 mol of N2O5 = 0.712 mol NO2 (0.356*2)
2. For the following reaction, calculate how many moles of each product are formed when 0.356
moles of PbS completely react. Assume there is an excess of oxygen.
2 PbS (s )+ 3O 2( g ) →2 Pb O(s )+2 SO 2(g )
Mole ratio: 2:3 -> 2:2
2PbO = 0.356 mol 2SO2 = 0.356 mol
3. For the following reaction, calculate how many grams of each product are formed when 4.05 g of
water is used.
2 H 2 O→ 2 H 2+O 2
2H2 = 0.907 g O2 = 3.584
4.05g H2O x (2mol H2O/36.032g H2O) = 0.225 mol H2O
0.225mol H2 x (2.016 H2/1mol H2) = 0.457 x 2 = .907 g H2
0.112mol O2 x (32g O2/1mol O2) = 3.584 g O2
4. Determine the theoretical yield of P2O5, when 3.07 g of P reacts with 6.09 g of oxygen in the
following chemical equation
4 P+5 O2 →2 P2 O5
7.034g P2O5
*Work attached to end of document.
5. Determine the percent yield of the following reaction when 2.80 g of P reacts with excess
oxygen. The actual yield of this reaction is determined to by 3.89 g of P 2O5.
, 4 P+5 O2 →2 P2 O5
Percent yield = 30.39%
*Work attached to end of document.
Properties of Liquids and Solids
Instructions: Answer the following questions using the following information: ∆Hfus=6.02 kJ/mol; ∆Hvap=
40.7 kJ/mol; specific heat of water is 4.184 J/g ∙˚C; specific heat of ice is 2.06 J/g∙˚C; specific heat of
water vapor is 2.03 J/g∙˚C.
1. How much heat is required to vaporize 25 g of water at 100˚C?
25g/18g = 1.389 moles of water
1.389 x 40.7 kJ = 56.532 kJ
2. How much heat is required to convert 25 g of ice at -4.0 ˚C to water vapor at 105 ˚C (report your
answer to three significant figures)?
75.8 kJ (work attached at bottom)
3. An ice cube at 0.00 ˚C with a mass of 8.32 g is placed into 55 g of water, initially at 25 ˚C. If no
heat is lost to the surroundings, what is the final temperature of the entire water sample after all
the ice is melted (report your answer to three significant figures)?
11.2°C (work attached at bottom)
Mixtures and Solutions
Instructions: Use the following image to answer the
following questions.
1. A solution contains 40 g of NaCl per 100 g of water
at 100˚C. Is the solution unsaturated, saturated or
supersaturated?
Saturated.
2. A solution contains 50 g of KCl per 100 g of water at
25˚C. Is the solution unsaturated, saturated or
supersaturated?
Supersaturated
3. A solution contains 10 g of KNO3 per 100 g of water at 30˚C. Is the solution unsaturated,
saturated or supersaturated?
Solubility Graph [Online Image]. Retrieved November 12,
2018 from https://socratic.org/chemistry/solutions-and-
their-behavior/solubility-graphs
Chemical Equations
Instructions: Solve the following chemical equations.
1. For the following reaction, calculate how many moles of NO 2 forms when 0.356 moles of the
reactant completely reacts.
2 N 2 O5( g) → 4 NO2( g )+O2( g)
2 moles of N2O5 for every 4 moles of NO2
0.356 mol of N2O5 = 0.712 mol NO2 (0.356*2)
2. For the following reaction, calculate how many moles of each product are formed when 0.356
moles of PbS completely react. Assume there is an excess of oxygen.
2 PbS (s )+ 3O 2( g ) →2 Pb O(s )+2 SO 2(g )
Mole ratio: 2:3 -> 2:2
2PbO = 0.356 mol 2SO2 = 0.356 mol
3. For the following reaction, calculate how many grams of each product are formed when 4.05 g of
water is used.
2 H 2 O→ 2 H 2+O 2
2H2 = 0.907 g O2 = 3.584
4.05g H2O x (2mol H2O/36.032g H2O) = 0.225 mol H2O
0.225mol H2 x (2.016 H2/1mol H2) = 0.457 x 2 = .907 g H2
0.112mol O2 x (32g O2/1mol O2) = 3.584 g O2
4. Determine the theoretical yield of P2O5, when 3.07 g of P reacts with 6.09 g of oxygen in the
following chemical equation
4 P+5 O2 →2 P2 O5
7.034g P2O5
*Work attached to end of document.
5. Determine the percent yield of the following reaction when 2.80 g of P reacts with excess
oxygen. The actual yield of this reaction is determined to by 3.89 g of P 2O5.
, 4 P+5 O2 →2 P2 O5
Percent yield = 30.39%
*Work attached to end of document.
Properties of Liquids and Solids
Instructions: Answer the following questions using the following information: ∆Hfus=6.02 kJ/mol; ∆Hvap=
40.7 kJ/mol; specific heat of water is 4.184 J/g ∙˚C; specific heat of ice is 2.06 J/g∙˚C; specific heat of
water vapor is 2.03 J/g∙˚C.
1. How much heat is required to vaporize 25 g of water at 100˚C?
25g/18g = 1.389 moles of water
1.389 x 40.7 kJ = 56.532 kJ
2. How much heat is required to convert 25 g of ice at -4.0 ˚C to water vapor at 105 ˚C (report your
answer to three significant figures)?
75.8 kJ (work attached at bottom)
3. An ice cube at 0.00 ˚C with a mass of 8.32 g is placed into 55 g of water, initially at 25 ˚C. If no
heat is lost to the surroundings, what is the final temperature of the entire water sample after all
the ice is melted (report your answer to three significant figures)?
11.2°C (work attached at bottom)
Mixtures and Solutions
Instructions: Use the following image to answer the
following questions.
1. A solution contains 40 g of NaCl per 100 g of water
at 100˚C. Is the solution unsaturated, saturated or
supersaturated?
Saturated.
2. A solution contains 50 g of KCl per 100 g of water at
25˚C. Is the solution unsaturated, saturated or
supersaturated?
Supersaturated
3. A solution contains 10 g of KNO3 per 100 g of water at 30˚C. Is the solution unsaturated,
saturated or supersaturated?
Solubility Graph [Online Image]. Retrieved November 12,
2018 from https://socratic.org/chemistry/solutions-and-
their-behavior/solubility-graphs
