HONS CHEMISTRY - CHAPTER 2 STOICHIOMETRY STOICHIOMETRY PROBLEMS WKST - ANS - V2 NAME: DATE: PAGE:. 200. g flour (s) + 40 eggs (s) + 5.0 L milk (L) 80 chocolate chip cookies (s) mass (g)? volume (L)? 00 cookies Compare the Recipe to the amounts of substances really used in the problem. 80 chocolate chip cookies 200. g flour = 00 chocolate chip cookies 80 = 2.00 0 4 g flour = 250. g flour 80 chocolate chip cookies 5.0 L milk = 00 chocolate chip cookies 80 = 5.0 0 2 L milk = 6.3 L milk
2. 2 C 6 H 6 (l) + 5 O 2 (g) 2 CO 2 (g) + 6 H 2 O (g) 0.0 mol mol? Compare the Recipe (The Balanced Equation) to the moles of substances really used in the problem. 2 mol C 6 H 6 = 0.0 mol C H 6 6 5 mol O 2 2 = 50. mol O 2 = 75.0 mol O 2 3. Al(OH) 3 (s) + 3 HCl (aq) AlCl 3 (aq) + mol? 5.00 mol 3 HOH (l) Compare the Recipe (The Balanced Equation) to the moles of substances really used in the problem. 3 mol HCl mol Al(OH = 5.00 mol HCl 3 = 5.00 mol Al(OH) 3 =.67 mol Al(OH) 3 2 - HC - Chapter 2 - Stoichiometry Problems Worksheet - Answers - V2
4. 2 O 2 (g) + CH 4 (g) CO 2 (g) + 2 H 2 O (g) 6.00 L volume (L)? All of the amounts in the Recipe are written in moles, which is great because all roads lead to moles! Convert the given amount to moles. 22.4 L O 2 = 6.00 L O 2 mol O 2 22.4 = 6.00 mol O 2 6.00 L O 2 mol O 2 = 0.268 mol O 2 22.4 L O 2 = 0.268 mol O 2 Compare the Recipe (The Balanced Equation) to the moles of substances really used in the problem. 2 mol O 2 = 0.268 mol O 2 mol CO 2 2 = 0.268 mol CO 2 = 0.34 mol CO 2 Convert the number of moles of CO 2 to volume (L). mol CO 2 = 0.34 mol CO 2 22.4 L CO 2 0.34 mol CO 2 22.4 L CO 2 = 3.00 L CO 2 mol CO 2 = 3.00 L CO 2 ~ ~ 3 - HC - Chapter 2 - Stoichiometry Problems Worksheet - Answers - V2
2 O 2 (g) + CH 4 (g) CO 2 (g) + 2 H 2 O (g) 6.00 L volume (L)? In Volume-Volume problems you always do the first mole conversion by dividing by 22.4 L and then do the second mole conversion by multiplying by 22.4 L. Why bother? Just use the Volume Ratio seen below. 2 volumes (L) O 2 = 6.00 L O 2 volume (L) CO 2 2 = 6.00 L = 3.00 L CO 2 5. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) 7.25 L volume (L)? Compare the Recipe (The Balanced Equation) in volume, to the volume of substances really used in the problem. 3 volumes (L) H 2 = 7.25 L H 2 2 volumes (L) NH 3 3 = 4.5 L = 4.83 L NH 3 4 - HC - Chapter 2 - Stoichiometry Problems Worksheet - Answers - V2
6. H 2 (g) + Cl 2 (g) 2 HCl (g) Volume (L)? 40.0 g All of the amounts in the Recipe are written in moles. This is great because all roads lead to moles! 36.5 g HCl mol HCl = 40.0 g HCl 36.5 = 40.0 mol HCl =.0 mol HCl 40.0 g HCl mol HCl =.0 moles HCl 36.5 g HCl Compare the Recipe (The Balanced Equation) to the moles of substances really used in the problem. 2 mol HCl mol H 2 =.0 moles HCl 2 =.0 mol H 2 = 0.548 mol H 2 Convert the number of moles of H 2 to volume (L). mol H 2 = 0.548 mol H 2 22.4 L H 2 0.548 mol H 2 22.4 L H 2 = 2.3 L H 2 mol H 2 = 2.3 L H 2 5 - HC - Chapter 2 - Stoichiometry Problems Worksheet - Answers - V2
7. You must now ask yourself (Self?): What is the balanced equation? If you passed the Chemical Reaction Test, then you of course know the answer. If you didn t pass, you must review and work on your techniques. Remember the book falls for everyone (Unless you live in outer space). But back to chemistry... C 6 H 2 (s) + 6 O 2 (g) 6 CO 2 (g) + 6 H 2 O (g) Mass (g)? 2.4 L Now we must convert the given volume into moles. So you ask yourself (Self?): What do I know about liters? If the drugs you take haven t corrupted your system (Yet), you will know that mol = 22.4 liters at STP. 22.4 L CO 2 = 2.4 L CO 2 mol CO 2 22.4 = 2.4 mol CO 2 2.4 L CO 2 mol CO 2 22.4 L CO 2 = 0.08 mol CO 2 = 0.08 mol CO 2 After you find the moles, the net step is to form a ratio Comparing The Recipe to the amount given in the problem. 6 mol CO 2 = 0.08 mol CO 2 mol C 6 H 2 6 = 0.08 mol C 6 H 2 = 0.079 mol C 6 H 2 Finally, you must find the Molar Mass of the glucose and convert the moles of glucose into mass (g). mol C 6 H 2 = 0.079 mol C H 2 6 80. g C 6 H 2 = 3.23 g C 6 H 2 0.079 mol C 6 H 2 80. g C 6 H 2 mol C 6 H 2 O 6 = 3.23 g C H 2 6 6 - HC - Chapter 2 - Stoichiometry Problems Worksheet - Answers - V2
8. 4 NH 3 (g) + 5 O 2 (g) 4 NO (g) + 6 H 2 O (g) mass (g)? 20. g Convert the given amount to moles. 32.0 g O 2 = 20. g O 2 mol O 2 32.0 = 20. mol O 2 20. g O 2 mol O 2 32.0 g O 2 = 3.75 mol O 2 = 3.75 mol O 2 Compare the Recipe (The Balanced Equation) to the moles of substances really used in the problem. 5 mol O 2 = 3.75 mol O 2 4 mol NH 3 5 = 5.0 mol NH 3 = 3.00 mol NH 3 Finally, you must find the Molar Mass of the NH 3 and convert the moles of NH 3 into mass (g). mol NH 3 = 3.00 mol NH 3 7.0 g NH 3 3.00 mol NH 3 7.0 grams NH 3 mol NH 3 = 5.0 grams NH 3 7 - HC - Chapter 2 - Stoichiometry Problems Worksheet - Answers - V2
= 5.0 g NH 3 8 - HC - Chapter 2 - Stoichiometry Problems Worksheet - Answers - V2
9. Al(NO 3 (aq) + 3 NaOH (aq) Al(OH (s) + 3 NaNO 3 (aq) Theoretical Mass 2.80 g? (g) Actual Mass of Product ----- 0.966 g We are looking for Percent Yield. We write the equation that contains that variable. Percent Yield = Actual Yield Theoretical Yield 00 ( ) The Actual Yield (Eperimental Yield) of Al(OH is given in the problem: 0.966 g Al(OH. Percent Yield = 0.966 g Al(OH Theoretical Yield 00 ( ) The Theoretical Yield (Calculated Yield or Epected Yield) is what we need to calculate, it is based on the 2.80 g Al(NO 3 reacting completely to produce the Al(OH. All of the amounts in the Recipe are written in moles. Convert the given amount to moles. 23 g Al(NO 3 = 2.80 g Al(NO 3 mol Al(NO 3 23 = 2.80 mol Al(NO 3 = 0.03 mol Al(NO 3 2.80 g Al(NO 3 mol Al(NO 3 = 0.03 mol Al(NO 3 23 g Al(NO 3 9 - HC - Chapter 2 - Stoichiometry Problems Worksheet - Answers - V2
Compare the Recipe (The Balanced Equation) to the moles of substances really used in the problem. mol Al(NO 3 = 0.03 mol Al(NO 3 mol Al(OH = 0.03 mol Al(OH) 3 Convert the number of moles of Al(OH to Mass (g). This mass will be the Theoretical Mass. mol Al(OH = 0.03 mol Al(OH 78.0 g Al(OH =.03 g Al(OH 0.03 mol Al(OH 78.0 g Al(OH =.03 g Al(OH mol Al(OH The Theoretical Yield (Calculated Yield or Epected Yield) is what we just calculated:.03 g Al(OH. We can now substitute this value in the equation so we can calculate the Percent Yield. Percent Yield = 0.966 g Al(OH.03 g Al(OH) 3 00 ( ) % Yield = 94.2% 0 - HC - Chapter 2 - Stoichiometry Problems Worksheet - Answers - V2