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otez555 [7]
3 years ago
15

Here is the balanced chemical equation for the combustion of isopropyl alcohol: 2C3H7OH(l) + 9O2(g) → 6CO2(g) + 8H2O(g) Multiply

ing 42 by 2/3, or 0.667, would be most useful for which calculation?
A. Calculating the moles of O2 needed to produce 42 moles of CO2.

B. Calculating the moles of CO2 that will form from 42 moles of C3H7OH.

C. Calculating the moles of O2 that will react completely with 42 moles of C3H7OH.

D. Calculating the moles of CO2 that will form from 42 moles of O2.
Chemistry
2 answers:
Whitepunk [10]3 years ago
5 0
I believe the correct answer from the choices listed above is option A. Multiplying 42 by 2/3, or 0.667, would be most useful for c<span>alculating the moles of O2 needed to produce 42 moles of CO2. Hope this answers the question.</span>
Lelechka [254]3 years ago
3 0
Hello there.

<span>Here is the balanced chemical equation for the combustion of isopropyl alcohol: 2C3H7OH(l) + 9O2(g) → 6CO2(g) + 8H2O(g) Multiplying 42 by 2/3, or 0.667, would be most useful for which calculation?

</span><span>A. Calculating the moles of O2 needed to produce 42 moles of CO2.
</span>
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Ca(OH)2 (s) precipitates when a 1.0 g sample of CaC2(s) is added to 1.0 L of distilled water at room temperature. If a 0.064 g s
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Answer:

D) Ca(OH)₂ will not precipitate because Q <  Ksp

Explanation:

Here we have first a chemical reaction in which Ca(OH)₂  is produced:

CaC₂(s)  + H₂O ⇒ Ca(OH)₂ + C₂H₂

Ca(OH)₂  is slightly soluble, and depending on its concentration it may precipitate out of solution.

The solubility product  constant for Ca(OH)₂  is:

Ca(OH)₂(s) ⇆ Ca²⁺(aq) + 2OH⁻(aq)

Ksp = [Ca²⁺][OH⁻]²

and the reaction quotient Q:

Q = [Ca²⁺][OH⁻]²

So by comparing Q with Ksp we will be able to determine if a precipitate will form.

From the stoichiometry of the reaction we know the number of moles of hydroxide produced, and since the volume is 1 L the molarity will also be known.

mol Ca(OH)₂ = mol CaC₂( reacted = 0.064 g / 64 g/mol = 0.001 mol Ca(OH)₂

the concentration of ions will be:

[Ca²⁺ ] = 0.001 mol / L 0.001 M

[OH⁻] = 2 x 0.001 M  = 0.002 M  ( From the coefficient 2 in the equilibrium)

Now we can calculate the reaction quotient.

Q=  [Ca²⁺][OH⁻]² = 0.001 x (0.002)² = 4.0 x 10⁻⁹

Q < Ksp since 4.0 x 10⁻⁹ < 8.0 x 10⁻⁸

Therefore no precipitate will form.

The answer that matches is option D

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3 years ago
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