You have two solutions, one 0.755 M barium nitrate and the other 1.250 M calcium hydroxide. (a)

You have two solutions, one 0.755 M barium nitrate and the...

You have two solutions, one $0.755 \mathrm{M}$ barium nitrate and the other $1.250 \mathrm{M}$ calcium hydroxide. (a) Write a net ionic equation for the precipitation reaction that occurs when these solutions are combined. (b) How many milliliters of the two solutions must be combined to prepare $5.00 \mathrm{~g}$ of precipitate? (c) Suppose you filter off the precipitate and find that your percent yield is $85.0 \%$. What volumes of the solutions should you have combined to isolate $5.00 \mathrm{~g}$ of precipitate?

You have two solutions, one $0.755 \mathrm{M}$ barium nitrate and the other $1.250 \mathrm{M}$ calcium hydroxide.
(a) Write a net ionic equation for the precipitation reaction that occurs when these solutions are combined.
(b) How many milliliters of the two solutions must be combined to prepare $5.00 \mathrm{~g}$ of precipitate?
(c) Suppose you filter off the precipitate and find that your percent yield is $85.0 \%$. What volumes of the solutions should you have combined to isolate $5.00 \mathrm{~g}$ of precipitate?

Key Concepts

Reaction Stoichiometry

Reaction stoichiometry involves the use of balanced chemical equations to relate the moles of reactants and products. It allows for the calculation of the amounts needed to produce a desired quantity of product, accounting for the stoichiometric ratios dictated by the reaction.

Percent Yield

Percent yield is a measure of the efficiency of a chemical reaction, comparing the actual mass of product obtained to the theoretical mass expected based on stoichiometric calculations. It is critical for adjusting reagent quantities to account for real-world losses and inefficiencies in the reaction process.

Solubility Rules

Solubility rules are guidelines used to predict the solubility of ionic compounds in water, which in turn indicate whether a compound will remain in solution or form a precipitate. They are crucial in determining the reaction’s outcome in precipitation reactions.

Net Ionic Equation

A net ionic equation displays only the ions and molecules directly involved in the chemical change, eliminating spectator ions that do not participate in the formation of the precipitate. This clarity simplifies the understanding of the actual chemical interaction occurring during the reaction.

Molarity and Volume Calculations

Molarity, defined as the number of moles of solute per liter of solution, is used to convert between the volume of a solution and the moles of its dissolved species. This concept is central to determining the exact volumes needed to provide the necessary amount of reactants for a reaction.

Transcript

00:01 If we have two solutions, one is 0 .755 molar of barium nitrate, and the other is 1 .25 molar of calcium hydroxide.

00:12 We have to first write the net ionic equation for the precipitation reaction that occurs, and then determine how many milliliters is needed for the precipitation to form 5 grams of the precipitate.

00:28 Our net ionic equation is just the equation or those ions that form the solid.

00:36 In this case, that solid is going to be barium hydroxide.

00:43 Let's write in our states.

00:45 I'm writing them up here just so that they are a little bit out of the way.

00:50 Now we have to figure out how many moles of this barium we have, but before we do that, we have to figure out the limiting reactant.

00:59 For barium, we're going to have 0 .755 molar, and for the hydroxide, it's 1 .250 times 2.

01:13 That's what this little 2 right up front means, so that becomes 2 .50 molar.

01:21 Now we have all of the information that we need to find our limiting reactant.

01:26 Right now, we have a 1 to 1 ratio with the barium.

01:32 If barium 2 +, we start out with 0 .755 molar of that, we would end up with exactly the same, 0 .755 molar.

01:48 Let's take a look at that hydroxide.

01:53 Here we have 2 .50 molar.

02:00 We're going to multiply that by some number of liters.

02:04 In this case, we'll just say 1 liter.

02:08 Then from here, we have to multiply it by our ratio of 1 to 2.

02:15 Then this just becomes 1 .250 molar.

02:20 In this case, barium is limiting because it produces a fewer number of moles.

02:28 Let's erase all of that, and let's start by figuring out how much barium is needed.

02:38 We're going to start with 5 grams because that's what we have, or that's what we want.

02:43 We're going to multiply it by the molar mass.

02:47 1 mole on the top of our fraction over 171 .34 grams on the bottom.

02:54 Now this is moles of barium hydroxide.

03:01 From here, we know 1 mole of our barium hydroxide comes from 1 mole of the barium solution.

03:18 We know so many x number of liters.

03:24 Well, that comes from 0 .755 moles.

03:33 Then from there, and that's of the barium solution, we say that 1 liter is equal to 1000 milliliters.

03:46 What we're going to do is we're going to figure out what that x is equal to.

03:52 Let's break out our calculators.

03:55 We're going to start with 5.

03:56 We're going to divide that by 171 .34.

04:00 We're going to multiply that by 1 over 1.

04:03 We're then going to divide it by 0 .755.

04:08 Now that x, that is 0 .03865.

04:16 If we multiply that by 1000, because we have to get it into milliliters, this becomes 38 .65 milliliters.

04:28 Now we can figure out how much hydroxide is needed.

04:33 We're going to start with 0 .755 molar of our barium.

04:42 Remember, this should make that a little bit of a bigger m there.

04:47 This big m, that stands for moles over liters.

04:54 Let's multiply it by our liters, 0 .03865 liters.

05:01 That gets us into moles.

05:05 Then we have our ratio of 2 moles of hydroxide is needed for every 1 mole of barium 2+.

05:19 Then from here, we have our some number of liters.

05:23 We don't know, but that's going to be divided by 2 .50 molar.

05:32 Then our last step is the 1000 milliliters over 1 liter.

05:39 That gets us into milliliters...

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