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Elenna [48]
3 years ago
10

5g of a mixture of KOH and KCl with water form a solution of 250mL. We have 25ml of this solution and we mix it with 14,3mL of H

Cl 0,1M. What's the % of the inicial mixture?
The 14,3mL of HCl are necessary to neutralize the KOH.

What are the reactions and what do I need to calculate?
Chemistry
1 answer:
cricket20 [7]3 years ago
7 0
We know that the number of moles HCl in 14.3mL of 0.1M HCl can be found by multiplying the volume (in L) by the concentration (in M).
(0.0143L HCl)x(0.1M HCl)=0.00143 moles HCl

Since HCl reacts with KOH in a one to one molar ratio (KOH+HCl⇒H₂O+KCl), the number of moles HCl used to neutralize KOH is the number of moles KOH. Therefore the 25mL solution had to contain 0.00143mol KOH.

To find the mass of KOH in the original mixture you have to divide the number of moles of KOH by the 0.025L to find the molarity of the KOH solution..
(0.00143mol KOH)/(0.025L)=0.0572M KOH

Since the morality does not change when you take some of the solution away, we know that the 250mL solution also had a molarity of 0.0572.  That being said you can find the number of moles the mixture had by multiplying 0.0572M KOH by 0.250L to get the number of moles of KOH.
(0.0572M KOH)x(0.250L)=0.0143mol KOH

Now you can find the mass of the KOH by multiplying it by its molar mass of 56.1g/mol.
0.0143molx56.1g/mol=0.802g KOH

Finally you can calulate the percent KOH of the original mixture by dividing the mass of the KOH by 5g.
0.802g/5g=0.1604
the original mixture was 16% KOH

I hope this helps.

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Answer:

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Explanation:

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Help on part "c": The forensic technician at a crime scene has just prepared a luminol stock solution by adding 19.0g of luminol
Contact [7]

1. The molarity of the stock solution of luminol = 1,431 M

2. 0.12 moles of luminol is present in 2.00 L of the diluted spray

3. The volume of the stock solution (Part A) would contain the number of moles present in the diluted solution (Part B): 83.86 ml

<h3>Further explanation</h3>

Stoichiometry in Chemistry studies about chemical reactions mainly emphasizing quantitative, such as the calculation of volume, mass, amount, which is related to the number of ions, molecules, elements, etc.

In stoichiometry included :

  • 1. Relative atomic mass
  • 2. Relative molecular mass

is the relative atomic mass of the molecule

  • 3. mole

1 mole is the number of particles contained in a substance with the same number of atoms in 12 gr C-12

1 mole = 6.02.10²³ particles

While the number of moles can also be obtained by dividing the mass (in grams) by the relative mass of the element or the relative mass of the molecule

\large{\boxed{\bold{mol\:=\:\frac{grams}{ relative\:mass} }}}

Luminol (C₈H₇N₃O₂) is a substance used to detect traces of blood in the scene of a crime because it reacts with iron in the blood

a. a luminol stock solution by adding 19.0g of luminol into a total volume of 75.0mL of H2O.

So the molarity is

  • 1. Luminol mole

- the relative molecular mass of Luminol

= 8. C + 7.H + 3.N + 2.16

= 8.12 + 7.1 + 3.14 + 2.16

= 177 grams / mol

so the mole:

mol = gram / relative molecular mass

mole=\frac{19}{177}

mole = 0.1073

2. Molarity (M)

M = mole / volume

M\:=\:{\frac{ 0.1703 }{75.10^{-3} L}

M = 1,431

  • b. luminol concentration in a spray bottle 6.00 × 10⁻² M. So that in 2 L of solution, the number of moles is:

mole = M x volume

mole = 6.10⁻² x 2

mole = 0.12

  • c. Molarity of the stock solution (Part A) = 1,431 M

the number of moles present in the diluted solution (Part B) = 0.12

So the volume of the stock solution (Part A) would contain the number of moles present in the diluted solution (Part B) is:

volume = mol / M

volume\:=\:\frac{0.12}{1.431}

volume = 0.08386 L = 83.86 ml

<h3>Learn more</h3>

moles of water you can produce

brainly.com/question/1405182

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Keywords: mole, volume, molarity, Luminol, the relative molecular mass

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Explanation:

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