When solutions of aluminum sulfate and sodium hydroxide are mixed, a white gelatinous precipitat

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When solutions of aluminum sulfate and sodium hydroxide are...

Key Concepts

Net Ionic Equation

A net ionic equation is a simplified chemical equation that shows only the species that are directly involved in the chemical change, excluding spectator ions that do not participate. This approach highlights the essential transformation, such as the formation of an insoluble compound in a precipitation reaction.

Precipitation Reaction

A precipitation reaction is a type of chemical reaction in which two aqueous solutions are combined to form an insoluble solid, known as the precipitate. The reaction is predicted by solubility rules and is useful for separating ions in analytical chemistry.

Stoichiometry

Stoichiometry involves quantifying the relationships between the reactants and products in a balanced chemical equation. It uses the mole ratios derived from the equation to calculate the amounts of substances consumed or produced, which is essential for predicting yields and converting between mass, volume, and molar units.

Limiting Reactant

The limiting reactant is the substance in a chemical reaction that is completely used up first, thereby limiting the amount of product formed. Identifying the limiting reactant is critical in stoichiometry, as it determines the maximum yield of the reaction.

Concentration and Molarity

Molarity is a common measure of concentration, defined as the number of moles of a solute per liter of solution. It is an essential concept for preparing solutions, performing quantitative reactions, and calculating the final concentration of ions after mixing or reaction.

Volume Additivity

Volume additivity is the assumption that the total volume of a mixture is equal to the sum of the individual volumes of the solutions added. This simplification is often applied in reaction calculations to determine the final concentration of reactants or products in the combined solution.

Transcript

00:01 So we've got some aluminum sulfate and sodium hydroxide solutions.

00:04 And they're going to come together and we're going to make some sodium sulfate, which is aqueous because of the sodium ion.

00:13 So the other product must be our precipitate.

00:16 That's our aluminum hydroxide.

00:22 So we're going to go ahead and balance this up.

00:28 And then we're going to turn this into a net ionic equation.

00:31 The simplest way to do that is to simply write your precipitate and then write the ions that form it.

00:42 So in that aluminum, there's an hydroxide, there is an aluminum ion, and there are three hydroxide ions.

00:52 So there's your net ion equation for the formation of our precipitate.

00:56 Now you'll notice that the stoichiometry in our net ionic equation and in our complete molecular equation are a little different.

01:04 Some of the mole ratios get kind of hidden in our net ionic equation.

01:09 So i always tell my students, when you're asking, to write a net an arc equation, go ahead and do that.

01:14 But if you're going to do stochometry, use your complete balanced equation instead.

01:20 So this problem is a limiting reactive problem.

01:24 They've given us amounts from both aluminum sulfate and for the sodium hydroxide.

01:29 So what we're going to want to do is just do the problem twice and choose the smaller of the two answers.

01:34 So i'm going to start with my 2 .76 grams of my aluminum sulfate.

01:44 And then, in three, typical stoichiometry fashion, i'm going to change grams to moles, and then i'm going to change moles to moles, and then the last step i'm going to change moles to grams.

01:56 So i'm changing grams of my aluminum sulfate to moles of my aluminum sulfate.

02:06 And then i'm going to go ahead and do my mole ratio, which is changing my aluminum sulfate to what we're looking for, which is our aluminum hydroxide.

02:21 And then in this last step, we'll be changing our aluminum hydroxide to grams, moles to grams.

02:31 So we'll use the molar mass of our aluminum sulfate, which is quite large.

02:38 Our moral ratio, you can get that from our equation, right? it's going to be two, right, to one.

02:45 So two to one.

02:46 And then our molar mass of aluminum hydroxide is 78.

02:52 So if you do all that, you're going to get 1 .26 grams of aluminum hydroxide.

03:01 So if aluminum sulfate is our limiting reactant, then that's our answer.

03:05 But we don't know yet.

03:06 So we're going to have to start now with the information they gave us for the sodium hydroxide, which they gave us a molarities, a malarity and a volume.

03:15 So i'm going to start with the molarity.

03:18 Malarity times liters.

03:20 Don't forget to change it to liters.

03:22 We'll give you the moles.

03:24 So i've got my moles here of my sodium hydroxide.

03:31 Now i can go ahead and do my mole ratio, moles to moles.

03:36 And then again that last step where we're going to go ahead and change moles to grams.

03:40 So i'm changing moles of sodium hydroxide to what i'm looking for, which is aluminum hydroxide.

03:49 And then i'm going to change my moles of aluminum hydroxide to grams of aluminum hydroxide.

03:58 Again, we know this is 78, and our mole ratio here, we're going to look in our equation for the sodium hydroxide is a 6, right, and the 2.

04:12 So this is going to be 2 to 6.

04:15 And if we multiply all this out, we're going to get 0 .553 grams of aluminum hydroxide.

04:27 So if aluminum sulfate is our limiting reactant, we would make this much.

04:33 But if sodium hydroxides are limiting reactant, we make this much.

04:37 So you choose the smaller of the two.

04:38 You're going to run out of the sodium hydroxide before you're going to run out of the aluminum sulfates.

04:48 And then what we want to do when we're done here is find the molarity of whatever ion is an excess.

04:56 Okay...

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