chapter 4 - examples Flashcards
(11 cards)
Example: Write a balanced chemical equation for the reaction of propane (C3H8) with oxygen (O2) to form carbon dioxide (CO2) and water (H2O).
Question: What are the steps to balance the chemical equation for the reaction of propane with oxygen?
Answer:
Write the equation with the correct formulas: C3H8 + O2 → CO2 + H2O
Balance the carbon atoms: C3H8 + O2 → 3CO2 + H2O
Balance the hydrogen atoms: C3H8 + O2 → 3CO2 + 4H2O
Balance the oxygen atoms: C3H8 + 5O2 → 3CO2 + 4H2O
Check to ensure the smallest set of whole numbers is used.
Sample Problem 5.5: How many molecules are contained in 5.0 moles of carbon dioxide (CO2)?
Question: How do you calculate the number of molecules in a given number of moles of a substance?
Answer:
Identify the original quantity and the desired quantity: 5.0 mol of CO2 (original quantity) and number of molecules of CO2 (desired quantity).
Write out the conversion factors: 1 mol = 6.02 x 10^23 molecules
Set up and solve the problem: 5.0 mol x (6.02 x 10^23 molecules / 1 mol) = 3.01 x 10^24 molecules
Example: Calculate the formula weight for FeSO4.
Question: How do you calculate the formula weight of a compound?
Answer:
Write the correct formula and determine the number of atoms of each element from the subscripts: FeSO4 contains 1 Fe atom, 1 S atom, and 4 O atoms.
Multiply the number of atoms of each element by the atomic weight and add the results:
1 Fe atom x 55.85 amu = 55.85 amu
1 S atom x 32.07 amu = 32.07 amu
4 O atoms x 16.00 amu = 64.00 amu
Formula weight of FeSO4 = 55.85 + 32.07 + 64.00 = 151.92 amu
Sample Problem 5.9: How many moles are present in 100 g of aspirin (C9H8O4, molar mass 180.2 g/mol)?
Question: How do you convert grams of a substance to moles using its molar mass?
Answer:
Identify the original quantity and the desired quantity: 100 g of aspirin (original quantity) and moles of aspirin (desired quantity).
Write out the conversion factors: 1 mol = 180.2 g
Set up and solve the problem: 100 g x (1 mol / 180.2 g) = 0.555 mol
Sample Problem 5.10: How many molecules are in a 325-mg tablet of aspirin (C9H8O4, molar mass 180.2 g/mol)?
Question: How do you convert milligrams of a substance to the number of molecules?
Answer:
Identify the original and desired quantities: 325 mg aspirin (original quantity) and molecules of aspirin (desired quantity).
Convert mg to g: 325 mg x (1 g / 1000 mg) = 0.325 g
Convert g to moles: 0.325 g x (1 mol / 180.2 g) = 0.0018 mol
Convert moles to molecules: 0.0018 mol x (6.02 x 10^23 molecules / mol) = 1.09 x 10^21 molecules
Sample Problem 5.11: Using the balanced chemical equation, how many moles of CO are produced from 3.5 moles of C2H6?
Question: How do you use mole ratios from a balanced chemical equation to convert between moles of different substances?
Answer:
Identify the original and desired quantities: 3.5 mol C2H6 (original quantity) and moles of CO (desired quantity).
Write out the conversion factors: 2 C2H6 + 5 O2 → 2 CO + 6 H2O (2 mol CO / 2 mol C2H6) = 1
Set up and solve the problem: 3.5 mol C2H6 x (2 mol CO / 2 mol C2H6) = 3.5 mol CO
Example: Using the balanced equation, how many grams of O3 are formed from 9.0 mol of O2?
Question: How do you convert moles of reactant to grams of product using a balanced chemical equation?
Answer:
Convert the number of moles of reactant to the number of moles of product using a mole–mole conversion factor: 3 O2 → 2 O3 9.0 mol O2 x (2 mol O3 / 3 mol O2) = 6.0 mol O3
Convert the number of moles of product to the number of grams of product using the product’s molar mass: 6.0 mol O3 x (48.0 g O3 / 1 mol O3) = 290 g O3
Example: Ethanol (C2H6O, molar mass 46.1 g/mol) is synthesized by reacting ethylene (C2H4, molar mass 28.1 g/mol) with water. How many grams of ethanol are formed from 14 g of ethylene?
Question: How do you convert grams of reactant to grams of product using a balanced chemical equation?
Answer:
Convert grams of reactant to moles: 14 g C2H4 x (1 mol C2H4 / 28.1 g C2H4) = 0.498 mol C2H4
Use the mole ratio to find moles of product: C2H4 + H2O → C2H6O 0.498 mol C2H4 x (1 mol C2H6O / 1 mol C2H4) = 0.498 mol C2H6O
Convert moles of product to grams: 0.498 mol C2H6O x (46.1 g C2H6O / 1 mol C2H6O) = 23 g C2H6O
Sample Problem 5.14: If the reaction of ethylene with water to form ethanol has a calculated theoretical yield of 23 g of ethanol, what is the percent yield if only 15 g of ethanol are actually formed?
Question: How do you calculate the percent yield of a reaction?
Answer:
Identify the actual yield and theoretical yield: 15 g (actual yield) and 23 g (theoretical yield).
Calculate the percent yield: Percent yield = (actual yield / theoretical yield) x 100% Percent yield = (15 g / 23 g) x 100% = 65%
Sample Problem 5.18: Determine how much of one reactant is needed to react with a second reactant.
Question: How do you determine the limiting reactant in a chemical reaction using the number of molecules?
Answer:
Write out the balanced equation: 2 H2 + O2 → 2 H2O
Determine the mole ratio: 2 molecules H2 / 1 molecule O2
Calculate the number of molecules of the second reactant needed for complete reaction: 4 molecules H2 x (1 molecule O2 / 2 molecules H2) = 2 molecules O2
Analyze the outcomes:
If the amount of O2 present is less than 2 molecules, O2 is the limiting reagent.
If the amount of O2 present is more than 2 molecules, O2 is in excess.
Sample Problem 5.20: Using the balanced equation, determine the limiting reactant when 10.0 g of N2 (MM = 28.02 g/mol) react with 10.0 g of O2 (MM = 32.00 g/mol).
Question: How do you determine the limiting reactant in a chemical reaction using the number of grams?
Answer:
Convert the number of grams of each reactant into moles using the molar masses: 10.0 g N2 x (1 mol N2 / 28.02 g N2) = 0.357 mol N2 10.0 g O2 x (1 mol O2 / 32.00 g O2
