Assigned Substance: Aluminum A) Reaction: Aluminum chloride is mixed with hydrogen gas to produce hydrochloric acid and aluminum metal. If you are given 5 grams of hydrogen and 50 grams of aluminum chloride, how many grams of aluminum can be produced if the reaction is 35% efficient? Step 1: ____ AlCl3 + ____ H2 → _____ HCl + _____ Al Step 2: Given the amounts of two reactants, determine which one is the limiting reactant by using a magic box to converting grams of one into grams of the other. Remember to think in terms of money to decide which is the limiting reactant (LR). Step 3: Calculate how many grams of your assigned substance can be produced starting with the amount of LR (grams → grams). Step 4: Figure out how much would actually be produced using the percent yield equation.

Quick answer:

3.54 g Al will be produced in this reaction.

Expert Answers

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Step 1: Balance the chemical equation.

There are 3 Cl atoms in 1 molecule of AlCl3 (left side of the equation). Due to conservation, there must be at least 3 Cl atoms on the right side of the equation, so you’ll have at least 3 molecules HCl on the left—but that means you’d need 3 H atoms on the right side, and since you’re working with molecules of H2, the coefficient for HCl should be divisible by 2. The correct number of HCl molecules in this equation is 6, because it is divisible by 3 (allowing you to balance the Cl atoms) and by 2 (allowing you to balance the molecules of H2). Having 6 HCl means you need 2 AlCl3; now you have 6 Cl atoms on each side. To get 6 atoms of H on the left, you need 3 molecules of H2. The coefficient for Al on the right is 2. Here’s the balanced equation:

2 AlCl3 + 3 H2 -> 6 HCl + 2 Al

Step 2: Convert to moles so you can compare the amounts of each substance directly.

Start with 5 grams of hydrogen—looking at the equation, we know this means 5 grams of H2. The mass of hydrogen on the periodic table is given in grams per mole—each H2 molecule weighs 2.01568 g/mol; use this as a conversion factor.

(5 grams H2) / (2.01568 g/mol) = 2.480 mol H2.

Do the same for AlCl3, which has a mass of 133.34 g/mol:

(50 grams AlCl3) / (133.34 g/mol) = 0.375 mol AlCl3. The actual mol ratio here is 6.613 mol H2 : 1 mol AlCl3—it’s given by (2.480 mol H2) / (0.375 mol AlCl3). The theoretical ratio (from the equation) is (3 mol H2) / (2 mol AlCl3) = 1.5 mol H2 : 1 mol AlCl3. The actual ratio is greater than the theoretical one, meaning that—in our experiment—we have more H2 than the reaction requires. H2 is present in excess and AlCl3 is the limiting reagent.

Step 3: Stoichiometry.

Looking back at the equation, we know that reacting 2 molecules of AlCl3 will produce 2 Al atoms. This is a 1:1 ratio, so reacting 0.375 mol AlCl3 gives us 0.375 mol Al (use 26.982 g/mol—the mass of Al—as a conversion factor to get 10.12 g Al). 10.12 g Al is the theoretical yield.

Step 4: Actual yield.

The reaction is only 35% efficient, so even though we could make 10.12 g Al, we’re only going to get 35% of that.

10.12 g Al (0.35) = 3.54 g Al. This is the actual yield.

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