Finally, check that all of the elements, and the total charges, are balanced correctly. If there is a difference in total charge between the reactants and products, then the charge will also have to be balanced. (Instructions for balancing charge are found in Chapter 11.)
Applications of Stoichiometry
Once you’ve balanced the chemical equation, you then have a very valuable tool for solving many chemical reaction problems on the MCAT. Perhaps the most useful bit of information to glean from a balanced reaction is the mole ratios of reactants consumed to products generated. Furthermore, you can also generate the mole ratio of one reactant to another or one product to another. All these ratios can be generated by a comparison of the stoichiometric coefficients. In the example involving the formation of water, we now understand that for every one mole of hydrogen gas consumed, one mole of water can be produced, but that for every one mole of oxygen gas consumed, two moles of water can be produced. Furthermore, we see that, mole-to-mole, hydrogen gas is being consumed at a rate twice that of oxygen gas.
Stoichiometry problems usually involve at least a few unit conversions, so you must be careful when working through these types of problems to ensure that units cancel out appropriately, leaving you with the desired unit(s) of the answer choices. Pay close attention to the following problem, which demonstrates a clear and easy-to-follow method for keeping track of the numbers, the calculations, and the unit conversions.
MCAT Expertise
Because the MCAT is a critical-thinking test, we might see application and analysis questions related to limiting reactants or percent yields.
Example: How many grams of calcium chloride are needed to prepare 72 g of silver chloride according to the following equation?
CaCl2 (
Solution: Noting first that the equation is balanced, 1 mole of CaCl2 yields 2 moles of AgCl when it is reacted with 2 moles of AgNO3. The molar mass of CaCl2 is 110 g, and the molar mass of AgCl is 144 g.
Thus, 27.5 g of CaCl2 are needed to produce 72 g of AgCl.
MCAT Expertise
When the quantities of two reactants are given on the MCAT, we should expect to have to figure out which is the limiting reactant.
Limiting Reactant
If you recall some of the experiments that you ran in your general chemistry lab, you know that rarely did you ever use stoichiometric quantities of compounds. Remember that stoichiometric coefficients are usually whole numbers and refer to the number of moles of the particular reactants and products. Some common compounds, like CaCO3, have molar masses in excess of 100 g/mol, and using such quantities per student would make for a very, very expensive lab. Rarely, then, are reactants added in the exact stoichiometric proportions as shown in the balanced equation. As a result, in most reactions, one reactant will be used up (consumed) first. This reactant is known as the limiting reactant because it limits the amount of product that can be formed in the reaction. The reactant that remains after all the limiting reactant is used up is called the excess reactant.
For problems involving the determination of the limiting reactant, you must keep in mind two principles:
1. All comparisons of reactants must be done in units of moles. Gram-to-gram comparisons will be useless and maybe even misleading.
2. It is not the absolute mole quantities of the reactants that determine which reactant is the limiting reactant. Rather, the rate at which the reactants are consumed (the stoichiometric ratios of the reactants) combined with the absolute mole quantities determines which reactant is the limiting reactant.
Example: If 28 g of Fe react with 24 g of S to produce FeS, what would be the limiting reagent? How many grams of excess reagent would be present in the vessel at the end of the reaction?
The balanced equation is Fe + S
Solution: First, determine the number of moles for each reactant.
Since 1 mole of Fe is needed to react with 1 mole of S, and there are 0.5 moles Fe for every 0.75 moles S, the limiting reagent is Fe. Thus, 0.5 moles of Fe will react with 0.5 moles of S, leaving an excess of 0.25 moles of S in the vessel. The mass of the excess reagent will be
Yields
