In this question, you are first given the masses of both reactants used to start the reaction. To figure out what will be left over, we must first determine which species is the limiting reagent. First, determine the molecular weight of each of the reactants: Na2S = 78.05 g/mol; AgNO3 = 169.9 g/mol. We find that we are given 0.5 mol Na2S for the reaction and 0.6669 mol AgNO3. Because we need two molar equivalents of AgNO3 for every mole of Na2S, AgNO3 is the limiting reagent, and the correct answer choice will be in grams of Na2S. Next, determine how much of the Na 2S will be left over by figuring out how much will be used if it reacts with all of the AgNO3.
Use half as much Na2S as AgNO3:
[(1.000 mol Na2S)/(2.000 mol AgNO3)] × 0.6669 mol AgNO3 = 0.3334 mol Na2S
Then subtract this amount of reagent used from the total available:
0.5000 mol Na2S-0.3334 mol Na2S = 0.1666 mol excess Na2S
Finally, determine the mass that this represents:
0.1666 mol excess Na2S × 78.05 g/mol Na2S = 13.00 g Na2S
(A) and (D) are incorrect because there will be no remaining AgNO3. (C) is the mass in 0.33 mol of Na2S, and you would have gotten this if we mixed it up with 0.166 mol.
8. A
You’re told that you must begin with some given mass of KClO3, x grams. To use it to determine a mass of another product, we must convert it to moles. Thus far, you would have
which is equal to the number of moles of KClO3.
This first step eliminates (C).
Next, convert the number of moles of KClO3 to the number of moles of oxygen, according to the balanced equation presented in the question stem:
Putting both steps together, the equation thus far is
This second step eliminates (B) because it does not use the correct molar ratio between KClO3 and O2. The final step is to convert the number of moles of oxygen to a mass, in grams of O2. At this point in your equation, the number of moles is in the numerator, and you want the number of grams of oxygen in the numerator, so multiply your ratio by the molar mass of oxygen, as shown in (A).
9. C
In the reaction, there is a single displacement (II), with the silver in silver oxide being replaced by the aluminum to form aluminum oxide. This single-displacement reaction also necessitates a transfer of electrons in a reduction/oxidation reaction or “redox” reaction (III). Therefore, the correct answer is both II and III, (C). A double-displacement reaction (I) typically takes two compounds and causes two displacements, whereas only one occurs in this question. Combination reactions (IV) generally take two atoms or molecules and combine them to form one product, usually with more reactants than products, which also is not the case in this question.
10. D
Typically, single-displacement (or oxidation/reduction) reactions and double-displacement reactions both have the same number of reactants and products. For example, single-displacement reactions are often of the following form (M = metal 1; M’ = metal 2; A = anion):
M + M’A
In this type of reaction, M takes the place of M’ in combining with an anion. Typically, a process of oxidation and reduction of the involved metals enables this. Double-displacement reactions also tend to have the same number of reactants and products, represented by this type of reaction (C = cation 1; C’ = cation 2; A = anion 1; A’ = anion 2):
CA + C’A’
In this reaction, the two compounds essentially “swap” anions/cations, beginning and ending with two compounds.
Combination reactions typically have more reactants than products, represented by the following reaction:
A + B
11. D
A net ionic equation represents each of the ions comprising the compounds in the reactants and products as individual ions, instead of combining them as molecules. Thus, (A) is not a net ionic reaction. The term net means that the correct answer does not include any spectator ions (ions that do not participate in the reaction). In this reaction, nitrate (NO3-) remains unchanged. You can eliminate any answer choice that includes it, which leaves only (D).
12. A
This is a popular equation that you’ve probably seen before. What’s missing from it are the coefficients! This is an unbalanced equation. To get anywhere with this problem, you have to balance it first.
6CO2 + 6H2O + energy
The theoretical yield is the amount of product synthesized if the limiting reagent is completely used up. This question therefore asks how much glucose is produced if the limiting reagent is 30 grams of water. First, calculate the number of moles of water represented by 30 grams by dividing by its molecular weight (18.01 g/mol). You’ll have a lot of conversion factors, so wait until the end to multiply it all out.
Next, add conversion factors to find the equivalent number of moles of glucose.
Finally, convert moles of glucose into grams of glucose by multiplying by its molecular weight (180.2 grams / mol glucose).
After crossing out equal terms from the numerator and denominator, you are left with this:
Dividing 18 and 6 into our numerators will yield 50 grams of glucose.
13. A
What you are shown is a net ionic equation. To answer this question, work backward from the amount of product to the reactant with its spectator ions (in this case, sulfate).
First, for every two moles of FeSCN we create, we must react two moles of Fe3+. Therefore, you’re looking for the mass of iron sulfate that can provide two moles of iron.
Next, determine the charge on sulfate, which is -2 and iron, which in this case is +3. Therefore, iron sulfate must be present as follows:
Fe2(SO4)3
The molecular formula tells you that each mole of iron sulfate releases two moles of atomic iron. Therefore, you only need one mole of iron sulfate for this reaction, which means the molar mass of iron sulfate is the answer.
Chapter 5: Chemical Kinetics and Equilibrium