Math  /  Numbers & Operations

Question(83. Lead ions can be precipitated from solution with KCl according to the reaction: Pb2+(aq)+2KCl(aq)PbCl2(s)+2 K+(aq)\mathrm{Pb}^{2+}(a q)+2 \mathrm{KCl}(a q) \longrightarrow \mathrm{PbCl}_{2}(s)+2 \mathrm{~K}^{+}(a q)
When 28.5 g KCl is added to a solution containing 25.7 g Pb2+25.7 \mathrm{~g} \mathrm{~Pb}^{2+}, a PbCl2\mathrm{PbCl}_{2} precipitate forms. The precipitate is filtered and dried and found to have a mass of 29.4 g . Determine the limiting reactant, theoretical yield of PbCl2\mathrm{PbCl}_{2}, and percent yield for the reaction.

Studdy Solution

STEP 1

What is this asking? We're mixing two chemicals, KCl and Pb2+, which react to make PbCl2.
We need to figure out which chemical runs out first (the *limiting reactant*), how much PbCl2 we *should* get in theory (the *theoretical yield*), and how much we *actually* got compared to what we expected (the *percent yield*). Watch out! Don't mix up grams and moles!
We need to convert everything to moles to see how the chemicals react with each other.
Also, remember that the percent yield is *not* the same as the amount of product we got.

STEP 2

1. Calculate the moles of reactants
2. Determine the limiting reactant
3. Calculate the theoretical yield
4. Calculate the percent yield

STEP 3

Alright, let's **start** by converting the grams of KCl to moles!
We know that the molar mass of KCl is 74.55 g/mol\text{74.55 g/mol}.
We have 28.5 g\text{28.5 g} of KCl, so we **divide** the mass by the molar mass to get the number of moles:
28.5 g74.55 g/mol=0.382 mol KCl\frac{28.5 \text{ g}}{74.55 \text{ g/mol}} = 0.382 \text{ mol KCl}

STEP 4

Now, let's do the same thing for Pb2+.
Its molar mass is 207.2 g/mol\text{207.2 g/mol}, and we have 25.7 g\text{25.7 g}:
25.7 g207.2 g/mol=0.124 mol Pb2+\frac{25.7 \text{ g}}{207.2 \text{ g/mol}} = 0.124 \text{ mol Pb}^{2+}

STEP 5

The chemical equation tells us that we need *two* moles of KCl for every *one* mole of Pb2+.
Let's see how many moles of KCl we'd need to react with all of our Pb2+:
0.124 mol Pb2+2 mol KCl1 mol Pb2+=0.248 mol KCl0.124 \text{ mol Pb}^{2+} \cdot \frac{2 \text{ mol KCl}}{1 \text{ mol Pb}^{2+}} = 0.248 \text{ mol KCl}

STEP 6

We have 0.382 mol\text{0.382 mol} of KCl, which is *more* than the 0.248 mol\text{0.248 mol} we need.
This means KCl is in **excess**, and Pb2+ is our **limiting reactant**!
Pb2+ will run out first, and that will stop the reaction.

STEP 7

Now that we know Pb2+ is the **limiting reactant**, we can use it to figure out how much PbCl2 we *should* get.
The chemical equation tells us that one mole of Pb2+ makes one mole of PbCl2.
So, if we have 0.124 mol\text{0.124 mol} of Pb2+, we *should* get 0.124 mol\text{0.124 mol} of PbCl2.

STEP 8

Let's convert that to grams.
The molar mass of PbCl2 is 278.1 g/mol\text{278.1 g/mol}:
0.124 mol PbCl2278.1 g1 mol=34.5 g PbCl20.124 \text{ mol PbCl}_2 \cdot \frac{278.1 \text{ g}}{1 \text{ mol}} = 34.5 \text{ g PbCl}_2So, our **theoretical yield** is 34.5 g\text{34.5 g} of PbCl2.

STEP 9

We *actually* got 29.4 g\text{29.4 g} of PbCl2.
To calculate the **percent yield**, we **divide** the actual yield by the theoretical yield and **multiply** by 100%:
29.4 g34.5 g100%=85.2%\frac{29.4 \text{ g}}{34.5 \text{ g}} \cdot 100\% = 85.2\%

STEP 10

The **limiting reactant** is Pb2+.
The **theoretical yield** of PbCl2 is 34.5 g\text{34.5 g}.
The **percent yield** of the reaction is \text{85.2%}.

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