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sweet-ann [11.9K]
3 years ago
8

Imagine it took 300 mL of 0.1 M LiOH to reach the first equivalence point and an additional 300 mL of 0.1 M LiOH (600 mL total)

to reach the second equivalence point. How many moles of H2 were in the original acid solution
Chemistry
1 answer:
Sedaia [141]3 years ago
8 0

Answer:

0.03 mol H₂

Explanation:

In a diprotic acid titration, the first equivalence point relates to the equilibrium:

  • H₂A + OH⁻ ↔ HA⁻ + H₂O

And the second equivalence point to:

  • HA⁻ + OH⁻   ↔ A⁻² + H₂O

We can add those two equations and we're left with:

  • H₂A + 2OH ⁻ ↔ A⁻² + 2H₂O

So to calculate the moles of H₂ that were in the original acid solution <u>we use the total volume used</u> (in this case 600 mL).

600 mL ⇒ 600/1000 = 0.6 L

We <u>calculate the moles of LiOH</u>, using its molar concentration:

  • 0.1 M * 0.6 L = 0.06 mol LiOH

And now we <u>convert moles of LiOH (or OH⁻) to moles of H₂ </u>using the  stoichiometric ratio:

  • 0.06 mol LiOH * \frac{1molH_{2}A}{2molLiOH} = 0.03 mol H₂

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First a balanced reaction equation must be established:
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Now if mass of aluminum = 145 g
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Now the mole ratio of Al : O₂ based on the equation is  4 : 3  
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∴ if moles of Al = 4.83 moles
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