Aluminum and chlorine combine to form aluminum chloride....

Aluminum and chlorine combine to form aluminum chloride. $(9.2,9.3,9.4,9.5)$ $$ 2 \mathrm{Al}(s)+3 \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{AlCl}_{3}(s) $$ a. How many moles of $\mathrm{Cl}_{2}$ are needed to react with $4.50 \mathrm{~mol}$ of $\mathrm{Al}$ ? b. How many grams of $\mathrm{AlCl}_{3}$ are produced when $50.2 \mathrm{~g}$ of $\mathrm{Al}$ reacts? c. When $13.5 \mathrm{~g}$ of $\mathrm{Al}$ reacts with $8.00 \mathrm{~g}$ of $\mathrm{Cl}_{2}$, how many grams of $\mathrm{AlCl}_{3}$ can form? d. If $45.0 \mathrm{~g}$ of $\mathrm{Al}$ and $62.0 \mathrm{~g}$ of $\mathrm{Cl}_{2}$ are mixed, and $66.5 \mathrm{~g}$ of $\mathrm{AlCl}_{3}$ is actually obtained, what is the percent yield of $\mathrm{AlCl}_{3}$ for the reaction?

Aluminum and chlorine combine to form aluminum chloride. $(9.2,9.3,9.4,9.5)$
$$
2 \mathrm{Al}(s)+3 \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{AlCl}_{3}(s)
$$
a. How many moles of $\mathrm{Cl}_{2}$ are needed to react with $4.50 \mathrm{~mol}$ of $\mathrm{Al}$ ?
b. How many grams of $\mathrm{AlCl}_{3}$ are produced when $50.2 \mathrm{~g}$ of $\mathrm{Al}$ reacts?
c. When $13.5 \mathrm{~g}$ of $\mathrm{Al}$ reacts with $8.00 \mathrm{~g}$ of $\mathrm{Cl}_{2}$, how many grams of $\mathrm{AlCl}_{3}$ can form?
d. If $45.0 \mathrm{~g}$ of $\mathrm{Al}$ and $62.0 \mathrm{~g}$ of $\mathrm{Cl}_{2}$ are mixed, and $66.5 \mathrm{~g}$ of $\mathrm{AlCl}_{3}$ is actually obtained, what is the percent yield of $\mathrm{AlCl}_{3}$ for the reaction?

Key Concepts

Balanced Chemical Equations

These equations represent the chemical reaction using symbols and formulas of reactants and products. They must be balanced to obey the law of conservation of mass, meaning the number of atoms for each element is the same on both sides. This concept underpins stoichiometric calculations in chemistry by providing the basis for determining how reactants combine and form products.

Mole Ratios

Mole ratios are derived from the coefficients in a balanced chemical equation and are used to relate the quantities of reactants and products. Understanding mole ratios is essential for converting between moles, mass, and number of particles for substances involved in a chemical reaction.

Molar Mass and Conversion

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole. It is used to convert between the mass of a substance and the number of moles, which is critical for quantitative analysis in chemical reactions.

Limiting Reactant

The limiting reactant is the substance that is entirely consumed first in a chemical reaction, thus determining the maximum amount of product that can be formed. Identifying the limiting reactant is crucial in stoichiometry for calculating theoretical yields.

Percent Yield

Percent yield is a measure of the efficiency of a chemical reaction, calculated by comparing the actual amount of product obtained to the theoretical amount predicted by stoichiometry. It is expressed as a percentage and highlights potential losses in a reaction process.

Transcript

00:04 In this question, we have the formation of aluminum chloride from aluminum and chlorine gas to produce two moles of alcl3.

00:16 We're starting from 2 .70 grams of aluminum and 4 .05 grams of chlorine.

00:27 For a, we're asked to predict which is, which reactant is limiting.

00:32 Starting from 2 .70 grams of aluminum.

00:35 Let's convert to moles of aluminum, 27 .0 grams in one mole.

00:46 Stoicometry here, two moles of aluminum to two moles of aluminum to two moles of aluminum chloride.

00:57 This works out to 0 .10 moles of al -c -l -3, and 4 .05 grams of cl3.

01:10 Molar mass 71 grams per mole, 3 moles of cl2 to 2 moles of aluminum chloride.

01:29 This works out to 0 .0380 moles of aluminum chloride.

01:38 So we see that the lesser amount of aluminum chloride is based on chlorine, therefore cl2 is the limiting reaction.

01:53 Reactant.

01:59 For b, what mass of aluminum chloride can be produced? we know that we're producing 0 .0380 moles of aluminum chloride, and its mass is 133 .5 grams per mole.

02:22 This works out to 5 .07 grams of alcl3, which can be produced for c.

02:40 What mass of the excess reactant remains.

02:45 We're going to start with the limiting reactant.

02:52 4 .05 grams of cl2.

02:55 We'll find out how much aluminum needs to react with this...

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