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新托福考试辅导_More Complex Stoichiometric Calculations

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Molar mass (44.11) (32.00) (44.01) (18.02)
Balanced equation C3H8 + 5O2 3CO2 + 4H2O
Mole:mole 1 5 3 4
No. of moles 2.18 10.9 6.54 8.72
Amount 96.1 g 349 g 146 L 5.251024
 
    You certainly don’t have to write out several tables; we just did that to make the method clearer to you. Once you practice, you won’t need to write the categories to the left, either. They will become second nature. In essence, you can work the problems faster and set yourself up nicely for a clear understanding of equilibrium problems in your future.
 
    Perhaps your teacher at school taught you to solve this type of problem using dimensional analysis. If so, and if you feel comfortable solving problems using that method, don’t learn to do it our way: stick to the method with which you feel comfortable. We’ll go through this problem using dimensional analysis: What mass of oxygen will react with 96.1 grams of propane? Again, you would first write the chemical formula and make sure your equation is correctly balanced:
C3H8 + 5O23CO2 + 4H2O
    The amount to start with, when setting up the dimensional analysis, is 96.1 g, and your goal is to calculate the number of grams of oxygen produced:
    Not too hard, right? Now, how many liters of CO2 would be produced at STP?
    And how many water molecules are produced?
    Some people prefer the table method, while others are more comfortable with dimensional analysis. Use whatever method you feel more comfortable using, but just be consistent—if you always do the same types of problems the same way, you’ll feel much more confident on test day. Now try the method you prefer on some problems.
 
    Examples
  1. Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide from the air; it reacts with carbon dioxide to form solid lithium carbonate and liquid water. What mass of gaseous carbon dioxide will be consumed in a reaction with 1.00 kg of lithium hydroxide?
        Explanation
     
        First write the reaction, then create and fill in your chart, and when you’re done filling in the necessary blocks, it should look like this:
     
    Molar mass (23.95) (44.01)
    Balanced equation 2LiOH + CO2 Li2CO3 + H2O
    No. moles 1000 g/23.95 g/mol = 41.8 moles 20.9 20.9 20.9
    Amount 1.00 kg 20.944.01 = 920 g
    1. Baking soda (NaHCO3) is often used as an antacid. It neutralizes excess hydrochloric acid secreted by the stomach in the reaction below:
      NaHCO3(s) + HCl(aq)NaCl(aq) + H2O(l ) + CO2(aq)
          How many grams of NaHCO3 would be needed to completely react with 10.0 g of HCl?
       
          Explanation
       
          According to the balanced equation above, 1 mole of NaHCO3 reacts with 1 mole of HCl. First you need to calculate how many moles of HCl are in 10 grams of HCl. The formula weight of HCl is about 36 g/mol, so you set up the equation:

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