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
@Zale101
@Jhannybean
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
\(Mass= Density * Volume\) 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.
@sleepyjess makes sense so far?
Yes
So now I would do 69.9*1.78?
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. \(124.42 g *\large\frac{1~mol}{39.1 g}= 3.182 mol~K\)
Yes, @sleepyjess
Now, we still need to know how many grams of KCl, we were only solving for K.
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 \(KCl \rightarrow K+Cl_2\)
You know what's molar ratio between KCl and K?
No
there's 1 moles of K, and 1 moles of KCl in the equation, correct?
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.
Here's a better understanding of molar ratio http://www.occc.edu/kmbailey/Chem1115Tutorials/Molar_Ratios.htm
Ok, so there is 3.182 of each?
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.
Oh, ok
Now what's next?
the question is looking for grams of KCl and we already have the mols for KCl
We need to convert the mols into grams, correct?
grams/mol = molar mass/1 mol
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
237.25
yes
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