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The following study guides are linked to Anne McKenna's website (http://employees.csbsju.edu/amckenna/CH123), with permission by the author. Stoichiometry
I Balancing Chemical Equations Law of Conservation of Mass Stoichimetric Conversions Balancing EquationsAll chemical reactions must obey the Law of Conservation of Mass: no mass can be created or destroyed during the reaction. This means that the number and type of each atom must remain the same – no atoms can be created or destroyed in the process. Because of the Law of Conservation of Mass, all chemical equations must be balanced before any quantitative information can be obtained from them. Let’s take a look at a sample equation and the various meanings it might have. Magnesium metal burns in oxygen to produce solid magnesium oxide. One way of displaying this information might be: Magnesium + oxygen à magnesium oxide This is not very helpful because it does not tell us much about the reactants (on the left of the arrow) or the product (on the right). A better equation would use the formulas for the reactants and products and would indicate the phase of the substances (solid, liquid, gas, or aqueous solution). So, let’s rewrite the equation: Magnesium has the symbol Mg and it is a solid Mg(s) Oxygen is diatomic and is a gas O2(g) Magnesium oxide is a solid and has the formula MgO(s) Mg(s) + O2(g) à MgO(s) This is better than the word equation above, but notice that we don’t have equal numbers of atoms on each side of the equation. There are two oxygen atoms on the left side of the equation and only one on the right. So, we must adjust the equation so that there are equal numbers of each type of atom on both sides of the equation. This process is called equation balancing. So, how can we get equal numbers of oxygen atoms on both sides of the equation? Once the formulas of all of the substances are written correctly, we balance equations using coefficients. A coefficient is a number placed in front of the formula. Coefficients multiply all atoms in the formula. To balance this equation, we would place a coefficient of 2 in front of the formula of MgO. Mg(s) + O2(g) à 2MgO(s) Now our oxygen atoms are balanced, but look at the magnesium atoms. We must adjust the number of magnesium atoms by multiplying the formula of Mg by 2. 2Mg(s) + O2(g) à 2MgO(s) If we check, we see that there are equal numbers of all atoms on both sides of the equation – in other words, the equation is balanced. Let’s look at this process with another example; then we’ll discuss the meaning of a balanced equation Example: Balance the following equation: potassium + water --> potassium hydroxide + hydrogen Potassium is K and is a solid K(s) Water is H2O and is a liquid H2O(l) Potassium hydroxide is dissolved in water and has the formula KOH KOH(aq) Hydrogen is a diatomic gas H2(g) K(s) + H2O(l) à KOH(aq) + H2(g) Notice that there are 2 hydrogen atoms on the left side and three on the right. Also notice that the number of hydrogen atoms must be even because they are paired up in H2O. So, the first step would be to get an even number of hydrogen atoms on the right. To do that, multiply the formula of KOH by a coefficient of 2. K(s) + H2O(l) à 2KOH(aq) + H2(g) Now, there are 2 hydrogen atoms on the left and 4 on the right, so multiply the formula of H2O by 2. K(s) +2 H2O(l) à 2KOH(aq) + H2(g) Finally, adjust the number of K atoms. 2K(s) +2 H2O(l) à 2KOH(aq) + H2(g) Done!!!! 1. Write balanced equations for the following:
What do balanced equations mean?Now that you can write balanced equations, let’s discuss how we can use them. Balanced equations give us information about the quantities (how much) of substances in chemical reactions. Let’s look at the equation describing the reaction of potassium with water that we balanced on the previous page. What meanings could this equation have? 2K(s) + 2 H2O(l) à 2KOH(aq) + H2(g) 2 atoms 2 molecules 2 formula units 1 molecule This equation tells us that in the lab we could combine 2 atoms of potassium with 2 molecules of water and we would produce 2 formula units of KOH and one molecule of hydrogen gas. Of course, we can’t manipulate individual atoms in the lab so let’s translate this reaction into a lab scale. Remember, it is more convenient for us to deal with groups of atoms in the lab, and that the grouping we use for atoms is the mole. So, let’s translate the equation into moles. 2K(s) + 2 H2O(l) à 2KOH(aq) + H2(g) 2 moles 2 moles 2 moles 1 mole If we react 2 moles of potassium metal with 2 moles of water, we can produce 2 moles of KOH and 1 mole of hydrogen gas. In all of the problems we do, we will use the coefficients obtained from balanced equations to represent moles of reactant or product. Using Balanced Equations to Predict Amounts of Product or Reactant The coefficients in balanced equations give us the simplest ratio of moles of reactants and products. We must use these mole ratios in order to convert from one substance to another in a problem. One way to remember this is to use the following stoichiometry flow chart:
In other words, in order to move from one substance to another, you must be in the units of moles. Let’s look at a few examples: Using the balanced equation for the reaction of potassium with water, how many moles of KOH and H2 can be produced by the reaction of 6 moles of K with an unlimited supply of water? We always start with a balanced equation. Always check the equation and balance it first. 2K(s) + 2 H2O(l) à 2KOH(aq) + H2(g) Place the amounts given and asked for above the formulas in the equation. 6 moles excess ? moles ? moles 2K(s) + 2 H2O(l) à 2KOH(aq) + H2(g) Remember that we must be in the units of moles to move from one substance to another (not a problem here). Use the mole ratios as conversion factors to move from one substance to another.
6 moles excess 6 moles 3 moles 2K(s) + 2 H2O(l) à 2KOH(aq) + H2(g) 2. How many moles of CO2 can be produced from the reaction of 5 moles of C4H10 with excess oxygen? Hint: use balanced equation from question 2b. The last time I checked, the lab did not contain any “mole measuring devices.” In the lab, we typically use grams (of a solid) or milliliters (of a solution or liquid) to measure amounts and then calculate a number of moles. For the time being, let’s concentrate on measuring in grams. As you learned previously, it is possible to convert from grams to moles by using the molar mass of a substance. 3. How many moles are represented by 42.32 grams of water? Let’s add the grams « moles step to the flow chart for stoichiometry:
Now we can go back to our potassium problem and begin with a number of grams of potassium. How many moles of hydrogen can be produced from 50.0 grams of potassium reacting with excess water? 50.0g excess ? mol 2K(s) + 2 H2O(l) à 2KOH(aq) + H2(g) Every time you need to convert from or calculate grams, you need a value for the molar mass. In this case, we need the molar mass of potassium (from the periodic table, 39.1 grams). We will place the conversion factor below the symbol in the equation. 50.0 g excess ? moles 2K(s) + 2 H2O(l) à 2KOH(aq) + H2(g)
1 mol= 39.1 g Now, calculate the number of moles of potassium and place the answer below the conversion factor.
50.0 g excess ? moles 2K(s) + 2 H2O(l) à 2KOH(aq) + H2(g)
1 mol= 39.1 g 1.28 mol If we have 1.28 mol of K, how many moles of H2 can we produce?
50.0 g excess ? moles 2K(s) + 2 H2O(l) à 2KOH(aq) + H2(g)
1 mol= 39.1 g 1.28 mol 0.640 mol 4. How many moles of carbon dioxide can be produced from the reaction of 10.0 grams of C4H10 with excess oxygen (see equation from 2b)?
Finally, we get to the point where we can measure both substances with grams. Let’s work through the problem with the (now worn out) potassium example: How many grams of potassium are required to react with excess water to produce15.5 grams of hydrogen? First, notice that we are given grams of product and asked to calculate grams of reactant. Using the coefficients in the balanced equation, you can move from any substance (reactant or product) to any other substance. Let’s categorize the information using the format above. ? g excess 15.5 g 2K(s) + 2 H2O(l) à 2KOH(aq) + H2(g)
For every gram conversion, we need a molar mass. We already know that 1 mole of K = 39.1 grams. The other conversion we will need is for H2, where 1 mole = 2.02 grams. We will put those in the equation and work from there. ? g excess 15.5 g 2K(s) + 2 H2O(l) à 2KOH(aq) + H2(g)
1 mol=39.1 g
1 mol = 2.02 g Our first step will be to calculate the number of moles of hydrogen. Place the answer below hydrogen in the equation.
? g excess 15.5 g 2K(s) + 2 H2O(l) à 2KOH(aq) + H2(g)
1 mol=39.1 g
1 mol = 2.02 g 7.67 mol Now we need to calculate the number of moles of K needed. You can do the conversion as we did in the previous example, or just notice that we need 2x as many moles of K as H2. Either way, we need 15.3 mol of H2. ? g excess 15.5 g 2K(s) + 2 H2O(l) à 2KOH(aq) + H2(g)
1 mol=39.1 g
1 mol = 2.02 g 15.3 mol 7.67 mol Our last step is to convert the number of moles of K to a number of grams.
5. How many grams of C4H10 are required to produce 100.0 grams of CO2 (see equation from 2b). 6. How many grams of silver chloride can be produced from the reaction of 25.3 grams of calcium chloride reacting with excess silver nitrate. The other product of the reaction is calcium nitrate. (Hint: balance equation first). | ||||||||||||||||||||||||||||||
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