Chapter 11 Natural Approach to Chemistry

Chapter 11 Natural Approach to Chemistry
Stoichiometry Chapter 11 Natural Approach to Chemistry Assignments: /38cd, 39cd, 43acde, 40cd /46-48; /49,50ab,51 /59-62 /64 and 66

Learning Objectives Apply the mole concept and the law of conservation of mass to calculate quantities of chemicals participating in reactions. Important terms: stoichiometry, percent yield, actual yield, theoretical yield, limiting reactant

Chemical equations tell stories…
2CO(g) + O2(g) → 2CO2(g) … and stories can be put into different categories Nonfiction Science fiction Adventure Romance History Psychology Children’s literature … Synthesis / Decomposition Single / Double replacement Precipitate reaction Polymerization reaction

2CO(g) + O2(g) 2CO2(g) Chemical equations tell stories…
But what exactly do they tell us? 2CO(g) O2(g) CO2(g) They tell us what compounds we start with: Carbon monoxide (CO) gas Oxygen (O2) gas what compounds are formed: Carbon dioxide (CO2) gas

They tell us how much of each compound is involved
Chemical equations tell stories… What else do they tell us? 2CO(g) O2(g) CO2(g) 2 CO molecules 1 O2 molecules 2 CO2 molecules They tell us how much of each compound is involved stoichiometry: the study of the amounts of substances involved in a chemical reaction.

2CO(g) + O2(g) 2CO2(g) 2 CO molecules 2 dozen CO molecules
2 moles CO molecules 2 x (6.023 x 1023) CO molecules 2 CO2 molecules 2 dozen CO2 molecules 2 moles CO2 molecules 2 x (6.023 x 1023) CO2 molecules 1 O2 molecules 1 dozen O2 molecules 1 mole O2 molecules (1 x) x 1023 O2 molecules

Number of moles is not conserved
Is that okay? 2CO(g) O2(g) CO2(g) Number of moles is not conserved 2 moles CO molecules + 1 mole O2 molecules ≠ 2 moles CO2 molecules Yes, as long as the chemical equation is balanced! The coefficients are important!!!

2CO(g) + O2(g) 2CO2(g) These are important! 2 moles CO molecules
Coefficients 2CO(g) O2(g) CO2(g) 2 moles CO molecules 1 mole O2 molecules 2 moles CO2 molecules This chemical equation is balanced The coefficients are correct

Coefficients are important
1 bag cake mix + 3 eggs + ¼ cup oil cup water batch cupcakes Write as a ratio:

These are stoichiometric equivalents
Fermentation of sugar (glucose) into alcohol: C6H12O6(aq) C2H5OH(aq) + 2CO2(g) 1 mole glucose 2 moles ethanol 2 moles carbon dioxide Write as a ratio: These are stoichiometric equivalents

Fermentation of sugar (glucose) into alcohol:
C6H12O6(aq) C2H5OH(aq) + 2CO2(g) 1 mole glucose 2 moles ethanol 2 moles carbon dioxide Write as a ratio: mole ratio: a ratio comparison between substances in a balanced equation. It is obtained from the coefficients in the balanced equation.

Mole ratios Fermentation of sugar (glucose) into alcohol:
C6H12O6(aq) C2H5OH(aq) + 2CO2(g) 1 mole glucose 2 moles ethanol 2 moles carbon dioxide mole ratios for this chemical equation 11.1 Analyzing a Chemical Reaction

Mole ratios Consider the following equation: CO(g) + 2H2(g) CH3OH(l)
carbon monoxide hydrogen methanol If the reaction produces 5 moles of CH3OH, how many moles of H2 are consumed? Asked: moles of H2 5 moles CH3OH x 2 moles H = moles H2 1 mole CH3OH

5 mole Cl2 x 2 mole AlCl3 = 3.3 mole AlCl3 3 mole Cl2
A mixture of aluminum metal and chlorine gas reacts to form aluminum chloride (AlCl3): 2Al(s) + 3Cl2(g) → 2AlCl3(s). How many moles of aluminum chloride will form when 5 moles of chlorine gas react with excess aluminum metal? Asked: moles AlCl3 Given: moles Cl2 5 mole Cl2 x 2 mole AlCl3 = mole AlCl3 3 mole Cl2

Potassium + Hydrogen Phosphate 
Finish the reaction in symbols and balance… If moles of hydrogen phosphate are consumed in the above reaction, how many moles of hydrogen are produced?

mass  moles…. There is no scale that measures in moles!
How do you convert from moles to grams? By using the molar mass (g/mole) The mass of 1 mole of Al is not the same as the mass of 1 mole of Cl2. How do you convert from grams of Al to grams of Cl2? By using the molar mass (g/mole) and mole ratios

Process for calculating grams from grams given

If 45.0 g of calcium carbonate (CaCO3) decomposes in the reaction CaCO3(s) → CaO(s) + CO2(g), how many grams of CO2 are produced? Asked: grams of CO2 Given: grams of CaCO3 Relationships: mole ratios molar mass of CaCO3 = ( x 3) = g/mole molar mass of CO2 = ( x 2) = g/mole Strategy:

If 45.0 g of calcium carbonate (CaCO3) decomposes in the reaction CaCO3(s) → CaO(s) + CO2(g), how many grams of CO2 are produced? B B 0.45 mole CaC03 x 1 mole CO2 x g CO2 = g CO2 1 mole CaC mole C02

CHAPTER 11 Stoichiometry 11.2 Percent Yield and Concentration

+ In theory, all 100 kernels should have popped.
Did you do something wrong? + 100 kernels 82 popped 18 unpopped

No + In theory, all 100 kernels should have popped.
Did you do something wrong? No In real life (and in the lab) things are often not perfect + 100 kernels 82 popped 18 unpopped

Percent yield What you get to eat! + 100 kernels 82 popped 18 unpopped

actual yield: the amount obtained in the lab in an actual experiment.
theoretical yield: the expected amount produced if everything reacted completely.

Percent yield in the lab
Decomposition of baking soda: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) Heating

Decomposition of baking soda: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) 4.87 g 10.00 g measured experimentally Can you think of reasons why the final mass of Na2CO3 may not be accurate? (What could be sources of error?) There is usually some human error, like not measuring exact amounts carefully Maybe the heating time was not long enough; not all the Na2HCO3 reacted - Maybe Na2CO3 was not completely dry; some H2O(l) was measured too - CO2 is a gas and does not get measured

Decomposition of baking soda: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 4.87 g measured experimentally Let’s calculate the percent yield obtained in experiment calculated

Decomposition of baking soda: 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 4.87 g measured experimentally Let’s calculate the percent yield calculated

Answer: Mass of D = 6.306 g Na2CO3
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) This is a gram-to-gram conversion: 10.00g NaHCO mole NaHCO mole Na2CO3 1mole Na2CO3 = 84.01 g NaHCO mole NaHCO g Na2CO3 Answer: Mass of D = g Na2CO3

The actual yield (measured) is 4.87 g.
2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) 10.00 g 4.87 g 10.00 g moles moles 6.306 g For g of starting material (NaHCO3), the theoretical yield for Na2CO3 is g. The actual yield (measured) is 4.87 g.

2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)
For g of starting material (NaHCO3), the theoretical yield for Na2CO3 is g. The actual yield (measured) is 4.87 g.

Stoichiometry with solutions
Decomposition of baking soda: (We just looked at this.) 2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g) 10.00 g Convert to moles Reaction of solid zinc with hydrochloric acid: Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq) 50.0 mL of a 3.0 M solution Reactions in solution Convert to moles

A sample of zinc metal (Zn) reacts with 50. 0 mL of a 3
A sample of zinc metal (Zn) reacts with 50.0 mL of a 3.0 M solution of hydrochloric acid (HCl) according to: Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq). How many grams of hydrogen gas (H2) will be produced? Assume zinc metal is present in excess. Relationships: M = mole/L Mole ratio: 2 moles HCl ~ 1 mole H2 Molar mass of H2 = x 2 = 2.02 g/mole Asked: grams of H2 produced Given: mL of 3.0 M HCl Reacting with excess zinc Solve: 0.0500L HCl x 3.0mole HCl x 1 mol H2 x g H2 = g H2 L HCl mol HCl mol H2 Answer: 0.15 grams of H2 are produced

Reaction of solid zinc with hydrochloric acid:
Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq) 50.0 mL of a 3.0 M solution Convert molarity to moles Sometimes the concentration is written in mass percent Vinegar is 5% acetic acid by mass

Asked: grams of acetic acid in 120 mL of vinegar
Commercial vinegar is reported to be 5% acetic acid (C2H4O2) by mass. How many grams of acetic acid are in 120 mL of commercial vinegar? (Assume the density of vinegar is the same as pure water, 1.0 g/mL.) Asked: grams of acetic acid in 120 mL of vinegar Given: 120 mL of vinegar and 5% acetic acid by mass Relationships: 120 mL = 120 g, given a density of 1.0 g/mL Solve: Answer: 6.0 g of acetic acid.

Let’s Review: Obtained from the experiment
Calculate using molar masses and mole ratios

Assignments 11.2:

Limiting Reactants Ch 11.3

Suppose you want to make 2 ham & cheese sandwiches
Can you still make 2 ham & cheese sandwiches if you have 4 slices of bread, 4 slices of ham, and 1 slice of cheese? Limiting factor No, you are limited by the cheese! You can only get 1 ham & cheese sandwich.

Limiting reactant Excess reactant limiting reactant: the reactant that “runs out” first in a chemical reaction. excess reactant: the reactant that is remaining after the reaction is complete.

Steps for Determining the Limiting Reactant
Convert both reactant masses to moles. Multiply by the mole ratio from the balanced equation to find how much reactant is needed to use up all of the other reactant. Compare the amounts of reactants. Compare what you have available to what you need. This gives you the amount you have available to use. This gives you the amount you need to consume all of the reactant. If what you need is more than what you have, then this is the limiting reactant.

Sample Problem: 364/58. Iron can be produced from the following reaction: Fe2O3 + 2Al  2Fe + Al2O3 a. If g of Fe2O3 reacts with 30.0 g Al, which one will be used up first?

Sample Problem: 365/58. Iron can be produced from the following reaction: Fe2O3 + 2Al  2Fe + Al2O3 b. For this next step, use the smaller mole answer from a. to find the needed amount

How much Fe can be produced?
1.112 mole Al mole Fe g Fe = g Fe 2 mole Al mole Fe How much of the excessive reactant is remaining? Fe2O3 is the excessive reactant. mole Al  mole Fe2O3  mass Fe2O3 Have – used = excess 1.112 mole Al 1 mole Fe2O g Fe2O3 = gFe2O3 2 mole Al mole Fe2O3 100.0g – g = g remaining (excess)

Asgn: 364/59, 60

11.3 Assignment 361/31,34,58-62

Solving Stoichiometric Problems
Chapter 14.4

Section 11.1 Section 11.2 Section 11.3 Section 11.4
Analyzing a Chemical Reaction Section 11.2 Percent Yield and Concentration Section 11.3 Limiting Reactants Section 11.4 Solving Stoichiometric Problems Use what we’ve learned to answer these questions: - What is the limiting reactant? - What is the theoretical yield? What is the percent yield? - How much excess reactant is left? - How much reactant is used if it’s in a solution?

What is the limiting reactant?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: When 48.0 g of Li reacts with 46.5 g of N2, which reactant is the limiting reactant? Lithium is the only group 1 metal that is capable of reacting directly with nitrogen gas. 6Li(s) + N2(g) → 2Li3N(s) Solve: 1) Moles of each reactant? 2) Apply the mole ratio 3) Compare what we have with what we need

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: When 48.0 g of Li reacts with 46.5 g of N2, which reactant is the limiting reactant? Lithium is the only group 1 metal that is capable of reacting directly with nitrogen gas. 6Li(s) + N2(g) → 2Li3N(s) Asked: Limiting reactant Given: g of Li (have) 46.5 g of N2 (have) molar mass of Li = g/mole molar mass of N2 = g/mole mole ratio: 6 moles Li ~ 1 mole N2

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 48.0 g of Li reacts with 46.5 g of N2 6Li(s) + N2(g) → 2Li3N(s)

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 6Li(s) + N2(g) → 2Li3N(s) ? 2) Apply the mole ratio We have: 6.92 moles Li; 1.66 moles N2 How much N2 do we need to react with 6.92 moles Li? Do we have enough N2? Yes, we have more than enough N2. That means we will run out of Li before we run out of N2

What is the theoretical yield?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: How much lithium nitride (LiN3) can be produced from this reaction? 6Li(s) + N2(g) → 2Li3N(s)

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: How much lithium nitride (Li3N) can be produced from this reaction? 6Li(s) + N2(g) → 2Li3N(s) Asked: Amount of Li3N produced Given: Li is the limiting reactant 6.92 moles Li (have) Relationships: molar mass of Li3N = g/mole mole ratio: 6 moles Li ~ 2 moles Li3N From the last problem

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: How much lithium nitride (Li3N) can be produced from this reaction? 6Li(s) + N2(g) → 2Li3N(s) Asked: Amount of Li3N produced Given: Li is the limiting reactant 6.92 moles Li (have) Relationships: molar mass of Li3N = g/mole mole ratio: 6 moles Li ~ 2 moles Li3N Solve: 1) Find moles of Li3N 2) Convert moles to grams

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 6Li(s) + N2(g) → 2Li3N(s) Asked: Amount of Li3N produced Given: Li is the limiting reactant 6.92 moles Li (have) Relationships: molar mass of Li3N = g/mole mole ratio: 6 moles Li ~ 2 moles Li3N Solve: 1) Find moles of Li3N 2) Convert moles to grams

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 6Li(s) + N2(g) → 2Li3N(s) Asked: Amount of Li3N produced Given: Li is the limiting reactant 6.92 moles Li (have) Relationships: molar mass of Li3N = g/mole mole ratio: 6 moles Li ~ 2 moles Li3N Solve: 1) Find moles of Li3N 2) Convert moles to grams Asked: g of Li3N are produced

What is the percent yield?
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: Calculate the percent yield of an experiment that actually produced 62.5 g of Li3N. 6Li(s) + N2(g) → 2Li3N(s)

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: Calculate the percent yield of an experiment that actually produced 62.5 g of Li3N. 6Li(s) + N2(g) → 2Li3N(s) Asked: Percent yield Given: Theoretical yield: g Li3N Actual yield: 62.5 g Li3N Relationships: Solve: Use the percent yield formula From the last problem

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 6Li(s) + N2(g) → 2Li3N(s) Asked: Percent yield Given: Theoretical yield: g Li3N Actual yield: 62.5 g Li3N Relationships: Solve: Use the percent yield formula

Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 6Li(s) + N2(g) → 2Li3N(s) Asked: Percent yield Given: Theoretical yield: g Li3N Actual yield: 62.5 g Li3N Relationships: Solve: Use the percent yield formula Answer: The percent yield in this particular experiment is 77.7%

What is left? Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: How much of the excess reactant remains after the limiting reactant is completely consumed? 6Li(s) + N2(g) → 2Li3N(s)

What is left? Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: How much of the excess reactant remains after the limiting reactant is completely consumed? 6Li(s) + N2(g) → 2Li3N(s) Asked: Amount of excess reactant left Given: N2 is the excess reactant 1.15 moles N2 (need) 1.66 moles N2 (have) Relationships: Molar mass of N2 = g/mole Solve: 1) How many moles N2 remain? 2) Convert moles to grams

What is left? 6Li(s) + N2(g) → 2Li3N(s)
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 6Li(s) + N2(g) → 2Li3N(s) Asked: Amount of excess reactant left Given: N2 is the excess reactant 1.15 moles N2 (need) 1.66 moles N2 (have) Relationships: 28.01 g/mole N2 Solve: 1) How many moles N2 remain? 2) Convert moles to grams

What is left? 2) Convert moles to grams 6Li(s) + N2(g) → 2Li3N(s)
Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 6Li(s) + N2(g) → 2Li3N(s) Asked: Amount of excess reactant left Given: N2 is the excess reactant 1.15 moles N2 (need) 1.66 moles N2 (have) Relationships: 28.01 g/mole N2 Solve: 1) How many moles N2 remain? 2) Convert moles to grams

What is left? Lithium metal (Li) reacts directly with nitrogen gas (N2) to produce lithium nitride (Li3N) according to the reaction: 6Li(s) + N2(g) → 2Li3N(s) Asked: Amount of excess reactant left Given: N2 is the excess reactant 1.15 moles N2 (need) 1.66 moles N2 (have) Relationships: 28.01 g/mole N2 Solve: 1) How many moles N2 remain? 2) Convert moles to grams Answer: 14 g of N2 will remain at the end of the reaction.

Molar Mass Al203 = 2Al+3(0) = 2(27)+3(16) = 102 g/mole
365/64. Aluminum metal reacts with oxygen in the air to form a layer of aluminum oxide according to the equation: 4Al + 3O2 --> 2Al203(s) If 28.0 g of Al reacts with excess oxygen in the air, what mass of aluminum oxide is formed? Molar Mass Al203 = 2Al+3(0) = 2(27)+3(16) = 102 g/mole 28.0gAl 1mole Al 2 mole Al g Al = g Al203 27g Al mole Al mole Al203

365/64. Aluminum metal reacts with oxygen in the air to form a layer of aluminum oxide according to the equation: 4Al + 3O2 --> 2Al203(s) B How many moles of oxygen are consumed during this reaction? 28 g Al 1mole Al 3mole O2 = mole O2 27 g Al 4 mole Al

C. If 250 g Al203 of is formed, how much Al reacted?
365/64. Aluminum metal reacts with oxygen in the air to form a layer of aluminum oxide according to the equation: 4Al + 3O2 --> 2Al203(s) C. If 250 g Al203 of is formed, how much Al reacted? 250 g Al mole Al mole Al g Al 102 g Al mole Al mole Al Answer: g Al

5.00g eth. 1 mole eth. = 0.108 moles ethanol 46.1 g eth
365/66. Wine can spoil when ethanol is converted to acetic acid by oxidation: C2H5OH + O2 --> CH3COOH + H2O Determine the limiting reactant when 5.00 g of ethanol (C2H5OH) and 2.0 g of oxygen are sealed in a bottle. 2C + 6H + O = = 46.1 g/mole 5.00g eth. 1 mole eth. = moles ethanol 46.1 g eth 2.0 g oxygen 1 mole oxy. = moles oxygen 32.0g oxy. O2 is the limiting reactant. We converted reactants to moles individually.

365/66. Wine can spoil when ethanol is converted to acetic acid by oxidation: C2H5OH + O2 --> CH3COOH + H2O b. Calculate how much acetic acid will form in grams. Start with the oxygen because it is the limiting reactant. moles O2 1 mole CH3COOH g CH3COOH 1 mole O mole CH3COOH Answer: g Molar mass: 2C + 4H + 2(O) = = 60 g/mole

365/66. Wine can spoil when ethanol is converted to acetic acid by oxidation: C2H5OH + O2 --> CH3COOH + H2O c. Calculate the amount of excess reactant remaining in grams. Excess reactant is ethanol. Available less used: moles – moles = moles moles eth. 46 g eth. = g eth. 1 mole eth. Amount of excess ethanol is g

Reactants in solution CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)
The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. What is the concentration of CuSO4 in the water if g of CuS precipitate is formed? The reaction is: CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s)

Reactants in solution The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. What is the concentration of CuSO4 in the water if g of CuS precipitate is formed? The reaction is: CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) Asked: Concentration of CuSO4(aq) Given: 1.0 L of solution is tested 0.021 g CuS (formed) Relationships: Molar mass of CuS = g/mole Mole ratio: 1 mole CuSO4 ~ 1 mole CuS

The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. What is the concentration of CuSO4 in the water if g of CuS precipitate is formed? The reaction is: CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) Asked: Concentration of CuSO4(aq) Given: 1.0 L of solution is tested 0.021 g CuS (formed) Relationships: Molar mass of CuS = g/mole Mole ratio: 1 mole CuSO4 ~ 1 mole CuS Solve: 1) How many moles of CuS? 2) How many moles of CuSO4? 3) What is the concentration of CuSO4?

The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is: CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) Asked: Concentration of CuSO4(aq) Given: 1.0 L of solution is tested 0.021 g CuS (formed) Relationships: 95.61 g/mole CuS 1 mole CuSO4 ~ 1 mole CuS Solve: 1) How many moles of CuS? 2) How many moles of CuSO4? 3) What is the concentration of CuSO4?

The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is: CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) Asked: Concentration of CuSO4(aq) Given: 1.0 L of solution is tested 0.021 g CuS (formed) Relationships: 95.61 g/mole CuS 1 mole CuSO4 ~ 1 mole CuS Solve: 1) How many moles of CuS? 2) How many moles of CuSO4? 3) What is the concentration of CuSO4? Have:

The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is: CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) Asked: Concentration of CuSO4(aq) Given: 1.0 L of solution is tested 0.021 g CuS (formed) Relationships: 95.61 g/mole CuS 1 mole CuSO4 ~ 1 mole CuS Solve: 1) How many moles of CuS? 2) How many moles of CuSO4? 3) What is the concentration of CuSO4? Answer: The concentration of CuSO4 is 2.20 x 10-4 M.

The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is: CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) Discussion: If the legal limit is 5.0 x 10-4 M of Cu2+, is the industrial plant following the environmental guidelines? Answer: The concentration of CuSO4 is 2.20 x 10-4 M.

Reactants in solution CuSO4(aq) → Cu2+(aq) + SO42–(aq)
The concentration of Cu2+ ions [found as CuSO4(aq)] in the water discharged from an industrial plant is found by adding an excess of sodium sulfide (Na2S) solution to 1.0 L of the contaminated water. The reaction is: CuSO4(aq) + Na2S(aq) → Na2SO4(aq) + CuS(s) Discussion: If the legal limit is 5.0 x 10-4 M of Cu2+, is the industrial plant following the environmental guidelines? Yes, because 2.20 x 10-4 M is less than the legal limit. Answer: The concentration of CuSO4 is 2.20 x 10-4 M. CuSO4(aq) → Cu2+(aq) + SO42–(aq)

Chapter 11 Natural Approach to Chemistry

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