﻿ PROBLEMS WITH AN EXCESS OF ONE REACTANT OR A LIMITING REACTANT - Stoichiometry (Chemical Calculations) and the Mole Concept - REVIEW OF MAJOR TOPICS - SAT Subject Test Chemistry ﻿

## PART 2 ## REVIEW OF MAJOR TOPICS ### PROBLEMS WITH AN EXCESS OF ONE REACTANT OR A LIMITING REACTANT

It will not always be true that the amounts given in a particular problem are exactly in the proportion required for the reaction to use up all of the reactants. In other words, at times some of one reactant will be left over after the other has been used up. This is similar to the situation in which two eggs are required to mix with one cup of flour in a particular recipe, and you have four eggs and four cups of flour.

TIP The paragraph describes a practical example of a “limiting reactant.”

Since two eggs require only one cup of flour, four eggs can use only two cups of flour and two cups of flour will be left over.

TIP Remember this recipe analogy to solve limiting reactant questions!

A chemical equation is very much like a recipe. Consider the following equation:

Zn(s) + 2HCl(l) → ZnCl2(aq) + H2(g)

If you are given 65 grams of zinc and 65 grams of HCl, how many grams of hydrogen gas can be produced? Which reactant will be left over? How many grams of this reactant will not be consumed? The given quantities are above the equation, and the equation masses are given beneath. To solve for the grams of hydrogen gas, the limiting reactant must be determined.

2nd step.          Compare the given quantities with the equation requirements.

Knowing that it takes 65 grams of zinc to react with 73 grams of hydrochloric acid, it is reasonable to surmise that, since there are only 65 grams of hydrochloric acid, not all of the 65 grams of zinc can be used. The limiting factor will be the amount of hydrochloric acid.

3rd step.          Solve for the quantity of product that will be produced

using the amount of the limiting factor.

Now that we know that the limiting factor is the amount of hydrochloric acid (65 g), the equation can be solved. This means that the proportion is set up ignoring the 65 grams of zinc. Solving for x, we get = 1.78 grams or 1.8 grams of hydrogen produced. This gives the proportion: ﻿

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