What's a good idea for a limiting reactant project
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Answer:
In a titration of 35.00 mL of 0.737 M H₂SO₄, 62.4 mL of a 0.827 M KOH solution is required for neutralization.
Explanation:
The balanced reaction is
H₂SO₄ + 2 KOH ⇒ 2 H₂O + K₂SO₄
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) 1 mole of H₂SO₄ is neutralized with 2 moles of KOH.
The molarity M being the number of moles of solute that are dissolved in a given volume, expressed as:
in units of
then the number of moles can be calculated as:
number of moles= molarity* volume
You have acid H₂SO₄
- 35.00 mL= 0.035 L (being 1,000 mL= 1 L)
- Molarity= 0.737 M
Then:
number of moles= 0.737 M* 0.035 L
number of moles= 0.0258
So you must neutralize 0.0258 moles of H₂SO₄. Now you can apply the following rule of three: if by stoichiometry 1 mole of H₂SO₄ are neutralized with 2 moles of KOH, 0.0258 moles of H₂SO₄ are neutralized with how many moles of KOH?
moles of KOH= 0.0516
Then 0.0516 moles of KOH are needed. So you know:
- Molarity= 0.827 M
- number of moles= 0.0516
- volume=?
Replacing in the definition of molarity:
Solving:
volume=0.0624 L= 62.4 mL
<u><em>In a titration of 35.00 mL of 0.737 M H₂SO₄, 62.4 mL of a 0.827 M KOH solution is required for neutralization.</em></u>
1) If the arrangement is symmetrical and the arrows are of equal length the molecule is nonpolar
2) If the arrows are of different lengths and if they do not balance each other the molecule is polar
3) If the arrangement is asymmetrical the molecule is polar
Answer:
Explanation:
Hello!
In this case, in a dilution process, water is added to a solute in order to decrease its concentration but increase the volume of the solution. It means that if we have 20.0 mL of a 12.0-M solution of HCl and we want a 0.500-M solution, we need to apply the following formula considering that the moles remain unchanged:
Thus, solving for the final volume is:
So plugging in the values we obtain:
Now, since the initial volume of acid was 20.0 mL and the final volume is 480 mL, the added volume of distilled water is:
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