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AlexFokin [52]
2 years ago
13

Using the balanced equation for the next few questions: 4 Fe(s) + 30 (9) - 2Fe2O3(s)

Chemistry
1 answer:
stira [4]2 years ago
5 0

Answer:

0.075 moles of iron oxide would be produced by complete reaction of 0.15  moles of iron.

Explanation:

The balanced reaction is:

4 Fe + 3 O₂ → 2 Fe₂O₃

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:

  • Fe: 4 moles
  • O₂: 3 moles
  • Fe₂O₃: 2 moles

You can apply the following rule of three: if by stoichiometry 4 moles of Fe produce 2 moles of Fe₂O₃, 0.15 moles of Fe produce how many moles of Fe₂O₃?

moles of Fe_{2} O_{3} =\frac{0.15 moles of Fe*2 moles of Fe_{2} O_{3}  }{4 moles of Fe}

moles of Fe₂O₃= 0.075

<u><em>0.075 moles of iron oxide would be produced by complete reaction of 0.15  moles of iron.</em></u>

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Partial pressure=mole fraction×Pt
x=0.044÷44(maolarmass of CO2)×Pt
x=0.044÷(44)2×Pt
x=5×10^-4×Pt
x=5×10^-4×Pt
where Pt:1atm=760mmHg
xatm=750mmHg
750×1÷760=0.99
now;5×10^-4×099=4.95×10^-4.
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3 years ago
How many moles of water are produced from 13.35 mol of oxygen?
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<span>2H2 + O2 mc015-1.jpg 2H2O

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2 years ago
A 50.00-mL solution of 0.0350 M aniline ( Kb = 3.8 × 10–10) is titrated with a 0.0113 M solution of hydrochloric acid as the tit
nexus9112 [7]

Answer:

pH = 3.70

Explanation:

Moles of aniline in solution are:

0.0500L × (0.0350mol / 1L) = <em>1.750x10⁻³ mol aniline</em>

Aniline is in equilibrium with water, thus:

C₆H₅NH₂ + H₂O ⇄ C₆H₅NH₃⁺ + OH⁻ Kb = 3.8x10⁻¹⁰

HCl reacts with aniline thus:

HCl + C₆H₅NH₂ → C₆H₅NH₃⁺ + Cl⁻

At equivalence point, all aniline reacts producing  C₆H₅NH₃⁺.  C₆H₅NH₃⁺ has its own equilibrium with water thus:

C₆H₅NH₃⁺(aq) + H₂O(l) ⇄ C₆H₅NH₂(aq) + H₃O⁺(aq)

Where Ka is defined as:

Ka = [C₆H₅NH₂] [H₃O⁺] / [C₆H₅NH₃⁺] = Kw / Kb = 1.0x10⁻¹⁴ / 3.8x10⁻¹⁰ = <em>2.63x10⁻⁵ </em><em>(1)</em>

<em />

As all aniline reacts producing C₆H₅NH₃⁺, moles in equilibrium are:

[C₆H₅NH₃⁺] = 1.750x10⁻³ mol - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in (1):

2.63x10⁻⁵ =  [X] [X] / [1.750x10⁻³ mol - X]

4.6x10⁻⁸ - 2.63x10⁻⁵X = X²

0 = X² + 2.63x10⁻⁵X - 4.6x10⁻⁸

Solving for X:

X = -2.3x10⁻⁴ → False answer, there is no negative concentrations

X = 2.0x10⁻⁴ → Right answer

As [H₃O⁺] = X, [H₃O⁺] = 2.0x10⁻⁴.

Now, pH = -log[H₃O⁺]. Thus, pH at equivalence point is:

<em>pH = 3.70</em>

<em></em>

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Answer:

B

Explanation:

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