[化学] More Complex Stoichiometric Calculations

When you’re asked to do stoichiometric calculations on the SAT IIChemistry exam, make sure that if you need to write out the chemicalformulas, you do this correctly. No matter how good you are at math andhow well you understand the stoichiometric rules that follow, you won’tget the right answer if your chemical formulas are wrong! If you feelthat you’re weak in this area, see the review (in Appendix II) ofchemical formula naming and writing. Perhaps the easiest way to approach problemsthat ask you to calculate the amounts of reactants consumed or productsproduced during the course of a reaction is to start by creating atable or chart. Let’s work through a typical example. Say the SAT IIChemistry test asks you what mass of oxygen will react completely with96.1 grams of propane. Notice that for this question, you’ll need tostart by writing the chemical formulas. Now follow these steps:

• Write the chemical equation.
• Calculate the molar masses and put them in parentheses abovethe formulas; soon you’ll figure out you don’t have to do this forevery reactant and product, just those you’re specifically asked about.
• Balance the equation.
• Next put any amounts that you were given into the table. Inthis example, you were told that the reaction started with 96.1 g ofpropane.
• Find the number of moles of any compounds for which you weregiven masses. Here you’d start with propane: you divide 96.1 grams bythe molar mass of propane (44.11 g/mol) to get the number of moles ofpropane (2.18 mol).
• Use the mole:mole ratio expressed in the coefficients ofeach of the compounds to find moles of all of the necessary compoundsinvolved. The only one you really need to know is oxygen, but let’s runthrough all of them for practice. If the coefficient for propane, whichis 1, is equal to 2.18 moles of propane, then the number of moles ofoxygen must be 52.18 = 10.9, the moles of CO2 is 32.18 = 6.54, and the moles of H2O = 42.18 = 8.72.

Molar mass	(44.11)	(32.00)	(44.01)	(18.02)
Balanced equation	C3H8 +	5O2	3CO2 +	4H2O
No. of moles	2.18	10.9	6.54	8.72
Amount	96.1 g

• Reread the problem to determine which amount was asked for. Thequestion asks for the mass of oxygen, so convert moles of oxygen tograms and you have the answer:

10.9 mol44.01 g/mol = 349 g oxygen

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

    But what if this question had asked you to determine the liters of CO2 consumed in this reaction at STP (273K, 1 atm)? You would take the number of moles of CO2 that we calculated from the table and use the standard molar volume for a gas, or 22.4 L/mol. So, 6.54 mol22.4 L/mol = 146 L.

    Finally, what if the question had asked howmany water molecules are produced? You would take the number of molesof water and multiply it by Avogadro’s number, 6.021023, to get 5.251024 molecules of water.

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 outseveral 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 theproblems faster and set yourself up nicely for a clear understanding of equilibrium problems in your future.

    Perhaps your teacher at school taught you tosolve this type of problem using dimensional analysis. If so, and ifyou feel comfortable solving problems using that method, don’t learn todo 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 upthe dimensional analysis, is 96.1 g, and your goal is to calculate thenumber 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, whileothers are more comfortable with dimensional analysis. Use whatevermethod you feel more comfortable using, but just be consistent—if youalways do the same types of problems the same way, you’ll feel muchmore confident on test day. Now try the method you prefer on someproblems.

    Examples

• Solid lithium hydroxide is used in space vehicles to remove exhaledcarbon dioxide from the air; it reacts with carbon dioxide to formsolid lithium carbonate and liquid water. What mass of gaseous carbondioxide will be consumed in a reaction with 1.00 kg of lithiumhydroxide?

    Explanation

    First write the reaction, then create and fillin your chart, and when you’re done filling in the necessary blocks, itshould 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

• 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 NaHCO3reacts with 1 mole of HCl. First you need to calculate how many molesof HCl are in 10 grams of HCl. The formula weight of HCl is about 36g/mol, so you set up the equation:

= 0.27 moles HCl.

    Now, knowing that 1 mole of baking sodareacts with 1 mole of NaCl, we next need to figure out how many gramsof baking soda would react with 0.27 moles of baking soda: 0.27 moles134 g/mol baking soda = 37.2 grams/mole NaHCO3

Molar mass	(85.32)	(36.46)
Balanced equation	NaHCO3 +	HCl	NaCl +	H2O +	CO2
No. moles	0.274	10.0/36.46 = 0.274
Amount	0.27485.32 = 23.4 g	10.0 g