What is the molarity of a solution containing 17.0 g of KCl in 213 mL of KCl solution?

Chemistry

1 answer:

Anarel [89]2 years ago

5 0

Answer:

1.070MKCl

Explanation:

So we know that the original formula is M= n/L (n being moles of solute, L being liters of solvent)

Since we do not have liters in this problem, we would need to convert milliliters to liters

213 mL= 0.213 L

We then see that we do not have moles, but we do have a mass, being 17.0 g. we would need to convert these grams to moles, giving us 0.228 mol.

Then, you would plug in 0.228 for your n, and now you are ready to solve your original formula, plugging everything in.

M=n/L

M=0.228 mol/0.213L

M= 1.070MKCl

I know this was long, but I hope this helps (:

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Answer: The balanced chemical equation is $2KClO_3(s)\rightarrow 2KCl(s)+3O_2(g)$

Explanation:

A balanced chemical equation is one where all the individual atoms are equal on both sides of the reaction. It follows the law of conservation of mass.

For the given unbalanced chemical equation:

$KClO_3(s)\rightarrow KCl(s)+O_2(g)$

On the reactant side:

Atoms of K = 1

Atoms of Cl = 1

Atoms of O = 3

On the product side:

Atoms of K = 1

Atoms of Cl = 1

Atoms of O = 2

To balance the equation, we must balance the atoms by adding 2 infront of both $KClO_3(s)$ and $KCl(s)$ . Also, a coefficient of 3 must be written infront of $O_2(g)$