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Stoichiometry Review Assignment #8

Presentation on theme: "Assignment 5.07: Solution Stoichiometry. Solution Concentration (Review of 5.06) Molarity = moles of solute/Liters of solution 25.2 g of NaCl dissolved."— Presentation transcript:

1 Assignment 5.07: Solution Stoichiometry

2 Solution Concentration (Review of 5.06) Molarity = moles of solute/Liters of solution 25.2 g of NaCl dissolved in 500. mL of Water. Percent by Mass = mass of solute/mass of solution 25.2 g of NaCl dissolved in 500. mL of Water.

3 Solution Stoichiometry (Sample) How many milliliters of 2.5 M Hydrochloric acid solution would be needed to react with 25.0 g of Iron (II) sulfide? FeS + 2HCl  H 2 S + FeCl 2 Step 1) 25.0 g FeS 1 mol FeS = 0.284 mol FeS 87.91 g FeS Step 2) 0.284 mol FeS 2 mol HCl = 0.569 mol HCl 1 mol FeS Step 3) 0.569 mol HCl 1 L Solution 1000 mL = 228 mL HCl 2.5 mol HCl 1 Lsolution

4 Solution Stoichiometry (Practice) How many milliliters of 3.5 M Hydrochloric acid solution would be needed to react completely with 32.0 g of Magnesium metal? Mg + 2HCl  MgCl 2 + H 2

5 Solution Stoichiometry (Sample) How many grams of chlorine gas are needed to react with 450. mL of a 1.5 M potassium bromide solution? Cl 2 + 2KBr → 2KCl + Br 2 Step 1) 450. mL KBr 1 L 1.5 mol KBr = 0.675 mol KBr 1000 mL 1 L solution Step 2) 0.675 mol KBr 1 mol Cl 2 = 0.338 mol Cl 2 2 mol KBr Step 3) 0.338 mol Cl 2 70.90 g Cl 2 = 23.9 g Cl 2 1 mol Cl 2

6 Solution Stoichiometry (Practice) How many grams of aluminum are needed to react with 745 mL of a 2.5 M iron (II) nitrate solution? 2Al (s) + 3Fe(NO 3 ) 2 (aq) → 3Fe (s) + 2Al(NO 3 ) 2 (aq)

7 Dilutions Stock solution - solutions with a high concentration, of commonly used aqueous solutions Dilute – decrease the concentration of a solution by adding more solvent

8 Dilutions (Sample) How would a student make 3.5 liters of a 2.50 M solution of acetic acid from a 12.0 M stock solution? Step 1) 3.5 L 2.50 mol = 8.75 mol Acetic Acid 1 L dilute sol. Step2) 8.75 mol 1 L stock sol. =.729 L stock sol. 12.0 mol Step 3) 3.5 liters -.729 liters = 2.77 liters of water added  Measure.729 L of the 12.0 M stock solution, then add 2.77 L of water to make 3.5 L of 2.5 M solution

9 Dilutions (Practice) How would a student make 100. mL of a 2.0 M solution of hydrochloric acid from a 12.0 M stock solution? How would a student make 100. mL of a 1.5 M solution from 18.0 M sulfuric acid?



Stoichiometry Review Assignment Answer Key. Example 1: Calculate the ... Example 3: How many atoms are in 1.67 moles of magnesium? 1.67 mol Mg . 6.02 x ...

Stoichiometry Review Assignment Answer Key Example 1: Calculate the mass of a magnesium, Mg, atoms in grams. 24.035 g Mg . 1 mol Mg . 1 molecule Mg = 4.04 x 10-23 g/Mg atom 1 mol Mg 6.02 x 1023 molecules 1 atom Mg Example 2: Calculate the number of atoms in one-millionth of a gram of magnesium, Mg. 1 x 10 -6 g Mg . 1 mol Mg . 6.02 x 1023 molecules . 1 atom Mg = 2.48 x 1016 atoms 24.30 g Mg 1 mol Mg 1 molecule Mg Example 3: How many atoms are in 1.67 moles of magnesium? 1.67 mol Mg . 6.02 x 1023 molecules . 1 atom Mg = 1.01 x 1024 atoms 1 mol Mg 1 molecule Mg Example 4: How many moles of magnesium are in 73.4 grams of magnesium? 73.4 g Mg . 1 mol Mg = 3.02 mol Mg 24.30 g Mg Example 7: Calculate the number of propane, C3H8 molecules, in 74.6 grams of propane. 74.6 g C3H8 . 1 mol C3H8 . 6.02 x 1023 molecules = 1.02 x 1024 molecules C3H8 1 mol C3H8 44 g C3H8 Example 8: What is the mass of 10.0 billion molecules of propane? 10 x 109 molecules C3H8 . 1 mol C3H8 . 44 g C3H8 = 7.31 x 10-13 g C3H8 6.02 x 1023 molecules 1 mol C3H8 Example 9: How many moles, molecules, and oxygen atoms are contained in 60-g sample of ozone O3? 1.25 mol ozone 60 g O3 . 1 mol O3 . = 48 g O3 1.25 mol O3 . 6.02 x 1023 molecules = 7.53 x 1023 molecules ozone = 22.5 x 1023 atoms O 1 mol ozone Example 13: What mass of phosphorous is contained in 45.3 grams of (NH4)3PO4? 1 mol P . 30.97 g P = 9.41 g P 45.3 g (NH4)3PO4 . 1 mol (NH4)3PO4 . 149.09 g (NH4)3PO4 1 mol (NH4)3PO4 1 mol P Example 14: What mass of ammonium phosphate would contain 15.0 g of nitrogen? 15.0 g N . 1 mol N . 1 mol (NH4)3PO4 . 149.09 g (NH4)3PO4 = 53.22 g P(NH4)3PO4 14 g N 3 mol N 1 mol (NH4)3PO4 Example 15: What mass of propane, C3H8, contains the same mass of carbon as is contained in 1.35 grams of barium carbonate, BaCO3? 1.35 g BaCO3. 1 mol BaCO3 . 1 mol C . 1 mol C3H8 . 44 g C3H8 = 0.097 g C3H8 149.09 g BaCO3 1 mol BaCO3 3 mol C 1 mol C3H8

Reaction StoichiometryAnswer Key C3H8 + 5O2 Æ 3CO2 + 4H2O Example 16: What mass of H2O is produced from the reaction of 6.3 g of propane? 6.3 g C3H8. 1 mol C3H8 . 4 mol H2O . 18 g H2O = 10.31 g H2O 44 g C3H8 1 mol C3H8 1 mol H2O Example 17: How many molecules of H2O are produced when 2 moles of O2 are reacted with excess propane? 2 mol O2 . 4 mol H2O . 6.02 x 1023 molecules = 9.63 x 1023 molecules H2O 1 mol H2O 5 mol O2 Example 18: How many molecules of H2O are produced when 10 molecules of oxygen react? 10 molecules O2 . 1 mol O2 . 4 mol H2O . 6.02 x 1023 molecules = 8 molecules H2O 6.02 x 1023 molecules 5 mol O2 1 mol H2O Example 19: How many atoms of hydrogen in water are produced when 2.4 moles of propane are reacted? 2.4 mol C3H8. 4 mol H2O . 2 mol H . 6.02 x 1023 atoms = 1.156 x 1025 atoms H2O 1 mol C3H8 1 mol H2O 1 mol H Example 20: How many molecules of CO2 are produced when 2.3 x 106 atoms of O2 are reacted? 2.3 x 106 atoms O2. 1 mol O2 . 6.02 x 1023 molecules . 3 mol CO2 . 6.02 x 1023 atoms = 6.9 x 105 atoms CO2 2 atoms O 1 mol O2 5 mol O2 1 mol CO2 Example 21: How many grams of O atoms in CO2 are produced when 23.2 g of propane are reacted? 23.2 g C3H8. 1 mol C3H8 . 3 mol CO2 . 2 mol O atoms . 16 g O = 50.62 g O 44 g C3H8 1 mol C3H8 1 mol CO2 1 mol O atom Example 22: How many grams of O in water are produced from the reaction of 11.2 grams of H in propane? 11.2 g H. 1 mol H . 1 mol C3H8 . 4 mol H2O . 1 mol O . 16 g O = 89.6 g O 1gH 8 mol H 1 mol C3H8 1 mol H2O 1 mol O Example 23: What is mass of CO2 produced when 6.5 g of propane is reacted with 14.2 g of O2? 6.5 g C3H8 . 1 mol C3H8 = 0.147 mol C3H8 . 5 mol O2 = 0.74 mol O2 44 g C3H8 1 mol C3H8 14.2 g O2 . 1 mol O2 = 0.44 mol O2 . 3 mol CO2 . 44 g CO2 = 11.7 g CO2 32 g O2 5 mol O2 1 mol CO2 Example 24: What is the number of atoms in H in H2O are produced when 2.9 x 1011 molecules of propane are reacted with 5.4 x 1012 molecules of O2? 2.9 x 1011 molecules C3H8 . 1 mol C3H8 = 4.82 x 10-13 mol C3H8 6.02 x 1023 molecules 12

5.4 x 10 molecules O2 .

1 mol O2 6.02 x 10

4.82 x 10

-13

23

= 8.97 x 10-12 mol O2

molecules

mol C3H8 . 4 mol H2O . 2 mol H . 6.02 x 1023 atoms = 2.32 x 1012 atoms H 1 mol C3H8 1 mol H2O 1 mol H

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