GRE Subject Test: Chemistry : Analyzing Reactions
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Example Questions
Example Question #11 : Acid Base Chemistry
250mL of 2N 
Basic and yellow
Acidic and green
Basic and green
Acidic and yellow
Neutral and green
Acidic and green
Each of these compounds requires one equivalent of H+ is added to 100mL of 5N NH3 mol, so for HClO4, 2N = 2M, and for NH3, 5N = 5M.
Using the concentrations and volumes, we can find the moles, finding 
In this case, equivalents of acid are equal to the equivalents of base, meaning that we are at the equivalence point in a titration. HClO4 is a strong acid, and NH3 is a weak base.
Thus, the acid will fully dissociate, while the base will not, resulting in a greater concentration of H+ than OH– in the solution. This means the resulting solution will be acidic. We know that the indicator changes from yellow to green at 8.3, which is a basic pH. Our initial solution is basic, and we must pass through the pH of 8.3 to reach our final acidic solution, with pH < 8.3, meaning that the indicator must change from yellow to green during the reaction. This gives out final answer that the solution will be acidic and green.
Example Question #11 : Reaction Calculations And Limiting Reagent
Given the unbalanced equation below, how many grams of carbon dioxide will be produced from one mole of glucose and three moles of oxygen?
The first step to solve will be to balance the chemical reaction:
We see that we see that for every one mole of glucose used, six moles of carbon dioxide will be made. Similarly, for every six moles of oxygen used, six moles of carbon dioxide will be formed. For the reaction to carry out to completion, however, there must exist six moles of oxygen for every one mole of glucose. In the problem's circumstances, one of these compounds becomes the limiting reactant, in this case it is oxygen.
We only have three moles of oxygen, but we would need six to react all the given glucose, making oxygen the limiting reagent. We need to find the carbon dioxide produced from the limited amount of oxygen present. Use the molar ratio between oxygen and carbon dioxide and the molar mass of carbon dioxide to solve.
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