Show:
Mass-Mole-Mass
Molarity/Titration
Limiting/Excess Reagent
Theoretical Yield
Volume-Mole-Volume
100
What is molar mass? What are moles?
Molar mass is the amount of grams a chemical has per 1 mole of it. Moles are units of measurement and used to find the number of molecules of an ANYTHING.
100
What is titration?
A common laboratory method of quantitative chemical analysis that is used to determine the unknown concentration of an identified analyte.
100
What is excess reagent? What is limiting reagent?
The excess reagent is left over reactants. The limiting reagent determines how far the reaction will go before the chemicals get used up and the reaction stops.
100
What is theoretical yield?
The amount of product obtained in a chemical reaction.
100
What is the unit that volume is measured in?
Liters
200
2 KClO3 ---> 2 KCl + 3 O2 1.50 mol of KClO3 decomposes. How many grams of O2 will be produced at STP?
1.50/x = 2/3 Cross multiply and divide 2.25 mol of O2 2.25 mol * 32g/mol = 72g.
200
How many grams of solute are present is 175mL of 6.7M glucose? (C6H12O6)
0.67M/L * 0.175L * 180g/mol = 21.11g.
200
For the combustion of sucrose: C12H22O11 + 12O2 ---> 12CO2 + 11H2O there are 10.0 g of sucrose and 10.0 g of oxygen reacting. Which is the limiting reagent?
1) Calculate moles of sucrose: 10.0 g / 342.2948 g/mol = 0.0292146 mol 2) Calculate moles of oxygen required to react with moles of sucrose: From the coefficients, we see that 12 moles of oxygen are require for every one mole of sucrose. Therefore: 0.0292146 mol times 12 = 0.3505752 mole of oxygen required 3) Determine limiting reagent: Oxygen on hand ⇒ 10.0 g / 31.9988 g/mol = 0.3125 mol Since the oxygen required is greater than that on hand, it will run out before the sucrose. Oxygen is the limiting reagent
200
2 FePO4 + 3 Na2SO4 = 1 Fe2(SO4)3 + 2 Na3PO4 If I perform this reaction with 25 grams of iron (III) phosphate and an excess of sodium sulfate, how many grams of iron (III) sulfate can I make?
33 grams
200
For gases at STP, one mole occupies a volume of 22.4 L. What volume will the following quantities of gases occupy? 1. 3.20 moles of O2 2. .750 moles of N2 3. 1.75 moles of CO2
1. 3.20 moles * 22.4 L = 71.68 L 2. .750 moles * 22.4 L = 16.8 L 3. 1.75 moles * 22.4 L = 39.2 L
300
To produce 2.75 mol of KCl. How many grams of KClO3 would be required? 2 KClO3 ---> 2 KCl + 3 O2
2.75 mol * 122.55 g/mol = 337 grams completes the task.
300
A 25 ml solution of 0.5 M NaOH is titrated until neutralized into a 50 ml sample of HCl. What was the concentration of the HCl?
Step 1 - Determine [OH-] Every mole of NaOH will have one mole of OH-. Therefore [OH-] = 0.5 M. Step 2 - Determine the number of moles of OH- Molarity = # of moles/volume # of moles = Molarity x Volume # of moles OH- = (0.5 M)(.025 L) # of moles OH- = 0.0125 mol Step 3 - Determine the number of moles of H+ When the base neutralizes the acid, the number of moles of H+ = the number of moles of OH-. Therefore the number of moles of H+ = 0.0125 moles. Step 4 - Determine concentration of HCl Every mole of HCl will produce one mole of H+, therefore number of moles of HCl = number of moles of H+. Molarity = # of moles/volume Molarity of HCl = (0.0125 mol)/(0.050 L) Molarity of HCl = 0.25 M The concentration of the HCl is 0.25 M.
300
Suppose 316.0 g aluminum sulfide reacts with 493.0 g of water. What mass of the excess reactant remains? The unbalanced equation is: Al2S3 + H2O ---> Al(OH)3 + H2S
1) Balance the equation: Al2S3 + 6H2O ---> 2Al(OH)3 + 3H2S 2) Determine moles, then limiting reagent: Al2S3 ⇒ 316.0 g / 150.159 g/mol = 2.104436 mol H2O ⇒ 493.0 g / 18.015 g/mol = 27.366 mol Al2S3 ⇒ 2.104436 / 1 = 2.104436 H2O ⇒ 27.366 / 6 = 4.561 Al2S3 is the limiting reagent. 3) Determine grams of water that react: The molar ratio to use is 1:6 1 is to 6 as 2.104436 mol is to x x = 12.626616 mol of water used 12.626616 mol times 18.105 g/mol = 227.4685 g 4) Determine excess: 493.0 g minus 227.46848724 = 265.5 g
300
If 18.5 grams of iron (III) sulfate are actually made when I do this reaction, what is my percent yield?
(18.5 / 33) x 100% = 56%
300
Calcium carbonate decomposes at high temperatures to form carbon dioxide and calcium oxide: CaCO3(s) = CO2(g) + CaO(s) How many grams of calcium carbonate will I need to form 3.45 liters of carbon dioxide?
14.1 grams
400
If 80.0 grams of O2 was produced, how many moles of KClO3 decomposed? 2 KClO3 ---> 2 KCl + 3 O2
1.67 mol of KClO3 decomposed.
400
What is the pH when 25.00 mL of 0.20 M CH3COOH has been titrated with 40.0 mL of 0.10 M NaOH?
pH = 5.354
400
Consider the reaction: 2H2 + O2 --> 2H2O Identify the limiting reagent in each of the reaction mixtures given below: a. 50 molecules of H2 and 25 molecules of O2. b. 0.8 mol H2 and 0.75 mol O2. c. 5 g H2 and 56 g O2.
a no reactant is in excess b H2 is the limiting reactant c Moles H2 = 5 g/ 2 = 2.5 moles O2 = 56 / 32 = 1.75 H2 is the limiting reactant ( 1.75 x 2 = 3.5 moles of H2 are needed)
400
If I do this reaction with 15 grams of sodium sulfate and get a 65.0% yield, how many grams of sodium phosphate will I make?
According to the stoichiometry, the theoretical yield is 11.5 grams. Multiplying this by 0.650, you get 7.48 grams.
400
Ethylene burns in oxygen to form carbon dioxide and water vapor: C2H4(g) + 3 O2(g) = 2 CO2(g) + 2 H2O(g) How many liters of water can be formed if 1.25 liters of ethylene are consumed in this reaction?
2.50 liters
500
2AgNO3 + Na2CrO4 = ag2CrO4 + 2NaNO3 A technician mixes a solution containing silver nitrate with a solution containing sodium chromate. 2.89 grams of precipitate is produced. Calculate the mass of silver nitrate present in the first solution.
Silver nitrate : Silver chromate = 2 : 1 Moles of Ag2CrO4 = 2.89g. / 332g/mol = 0.00871 moles Moles of 2AgNO3 = 0.00871 * 2 = 0.0174 moles Mass 2AgNO3 = 0.0174mol * 169.88g/mol = 2.95g.
500
Consider the titration of a 24.0-mL sample of 0.105 M CH3COOH with 0.130 M NaOH. What is a) the initial pH? b) the volume of added base required to reach the equivalence point? c) the pH at 6.00 mL of added base? d) the pH at one-half of the equivalence point? e) the pH at the equivalence point?
a) pH = 2.865 b) 19.4 mL c) pH = 4.404 d) pH = 4.752 e) pH = 9.994
500
aluminium can be used to reduce iron (III) oxide to iron: Fe203 + 2Al --> Al203 + 2Fe If 30.9g of iron (III) oxide is treated with 10.0g of aluminium a. what mass of the excess reagent would remain at the end of the reaction b. what mass of iron is formed
a. 1mol Fe2O3 reacts with 2mol Al Molar mass Fe2O3 =159.6887 g/mol 30.9g = 30.9/159.6887 = 0.1935mol This will require 2*0.1935mol Al = 0.387mol Al Molar Mass Al = 26.98g/mol 10g = 10/26.98 = 0.3706mol this is less than the required amount. Al is therefore limiting Calculate mass of excess Fe2O3: You have 0.3706mol Al. This will react with 0.3706/2 = 0.1853mol Fe2O3 You have started with 0.1935mol Fe2O3. Therefore mol Fe2O3 unreacted = 0.1935 - 0.1853 = 0.0082mol Fe2O3 unreacted Molar mass Fe2O3 = 159.6887g/mol 0.0082mol = 1.309 = 1.31 g Fe2O3 unreacted b. 1mol Al will produce 1 mol Fe 0.3706 mol Al will produce 0.3706 mol Fe Molar mass Fe = 55.847 0.3706mol = 55.847*0.3706 =20.699g Mass of Fe produced = 20.7g
500
A student adds 200.0g of C7H6O3 to an excess of C4H6O3, this produces C9H8O4 and C2H4O2. Calculate the percent yield if 231 g of aspirin (C9H8O4) is produced. C7H6O3 + C4H6O3 = C9H8O4 + C2H4O2
≈90%
500
When chlorine is added to acetylene, 1,1,2,2-tetrachloroethane is formed: 2 Cl2(g) + C2H2(g) = C2H2Cl4(l) How many liters of chlorine will be needed to make 75.0 grams of C2H2Cl4?
21.8 L