Scientific knowledge results from the interaction of scientists who are investigating natural phenomena. Which of these BEST des
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Answer:
- The molarity of the student's sodium hydroxide solution is 0.0219 M
Explanation:
<u>1) Chemical reaction.</u>
a) Kind of reaction: neutralization
b) General form: acid + base → salt + water
c) Word equation:
- sodium hydroxide + oxalic acid → sodium oxalate + water
d) Chemical equation:
- NaOH + H₂C₂O₄ → Na₂C₂O₄ + H₂O
b) Balanced chemical equation:
- 2NaOH + H₂C₂O₄ → Na₂C₂O₄ + 2H₂O
<u>2) Mole ratio</u>
- 2mol Na OH : 1 mol H₂C₂O₄ :1 mol Na₂C₂O₄ : 2 mol H₂O
<u>3) Starting amount of oxalic acid</u>
- mass = 28 mg = 0.028 g
- molar mass = 90.03 g/mol
- Convert mass in grams to number of moles, n:
n = mass in grams / molar mass = 0.028 g / 90.03 g/mol = 0.000311 mol
<u>4) Titration</u>
- Volume of base: 28.4 mL = 0.0248 liter
- Concentration of base: x (unknwon)
- Number of moles of acid: 2.52 mol (calculated above)
- Proportion using the theoretical mole ratio (2mol Na OH : 1 mol H₂C₂O₄)
That means that there are 0.000622 moles of NaOH (solute)
<u>5) Molarity of NaOH solution</u>
- M = n / V (liter) = 0.000622 mol / 0.0284 liter = 0.0219 M
That is the correct number using <em>three signficant figures</em>, such as the starting data are reported.
Taking into account the reaction stoichiometry, 2 moles of CaO are required to react with 2 moles of Ca(OH)₂.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
CaO + H₂O → Ca(OH)₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- CaO: 1 mole
- H₂O: 1 mole
- Ca(OH)₂: 1 mole
The following rule of three can be applied: If by stoichiometric reaction 1 mole of Ca(OH)₂ is produced by 1 mole of CaO, 2 moles of Ca(OH)₂ are produced by how many moles of CaO?
moles of CaO= 2 moles
Finally, 2 moles of CaO are required to react with 2 moles of Ca(OH)₂.
Learn more about the reaction stoichiometry:
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