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Chemistry 17 Online
OpenStudy (sleepyjess):

How many grams of potassium chloride must decompose to produce 69.9 milliliters of potassium metal, if the density of potassium is 1.78 g/mL? KCl → K + Cl2

OpenStudy (sleepyjess):

@Zale101

OpenStudy (sleepyjess):

@Jhannybean

OpenStudy (zale101):

We are looking for he mass of KCl which will decompose and produce 69.9 ml of K. The density of K is already given which is 1.78 g/ml. Let's find the mass of Potassium. We know that \(\large Density=\Large\frac{Mass}{Volume}\) , but we have the volume given which is 69.9 mL of K and density which is 1.78 g/mL That means we solve for the mass by myltiply volumes to both sides canceling the volume from the right side

OpenStudy (zale101):

\(Mass= Density * Volume\) Mass (K)= 69.9 mL*\(\Large\frac{1.78g}{1mL}\) the mL will canceling out and we'll have the answer in grams for Potassium which is K.

OpenStudy (zale101):

@sleepyjess makes sense so far?

OpenStudy (sleepyjess):

Yes

OpenStudy (sleepyjess):

So now I would do 69.9*1.78?

OpenStudy (zale101):

Mass of K=124.42 g Now, we need to convert this to moles. To do that, we divide the grams by the atomic weight of K which is 39.1g/mol. \(124.42 g *\large\frac{1~mol}{39.1 g}= 3.182 mol~K\)

OpenStudy (zale101):

Yes, @sleepyjess

OpenStudy (zale101):

Now, we still need to know how many grams of KCl, we were only solving for K.

OpenStudy (zale101):

We got the moles of K though which will help us know the moles of KCl if we looked at the molar ratio from the balanced chemical equation \(KCl \rightarrow K+Cl_2\)

OpenStudy (zale101):

You know what's molar ratio between KCl and K?

OpenStudy (sleepyjess):

No

OpenStudy (zale101):

there's 1 moles of K, and 1 moles of KCl in the equation, correct?

OpenStudy (zale101):

Molar ratio is 1 to 1 which tells us that there were moles of K produced that has the same amount of moles in KCl when decomposed . KCl has 3.182 mols as well.

OpenStudy (zale101):

Here's a better understanding of molar ratio http://www.occc.edu/kmbailey/Chem1115Tutorials/Molar_Ratios.htm

OpenStudy (sleepyjess):

Ok, so there is 3.182 of each?

OpenStudy (zale101):

We had 3.182 mols of KCl before decomposing and after decomposing we produced 3.182 mols of K. So K and KCl has the same amount of mols, yes, by molar ratio.

OpenStudy (sleepyjess):

Oh, ok

OpenStudy (zale101):

Now what's next?

OpenStudy (zale101):

the question is looking for grams of KCl and we already have the mols for KCl

OpenStudy (zale101):

We need to convert the mols into grams, correct?

OpenStudy (sleepyjess):

grams/mol = molar mass/1 mol

OpenStudy (zale101):

Yeah, \(grams~KCl=3.182~mol(KCl)\Large \frac{74.56 g}{mole}\small(KCl)\) Moles cancel out, that leaves us with multiplying 3.182 by 74.56 g to get the grams of KCl

OpenStudy (sleepyjess):

237.25

OpenStudy (zale101):

yes

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