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Mars2501 [29]
3 years ago
8

Consider the reaction 4FeS2 + 11O2 → 2Fe2O3 + 8SO2. If 8 moles of FeS2 react with 15 moles of O2, what is the limiting reactant?

(3 points)
Select one:

a. FeS2

b. Fe2O3

c. O2

d. SO2
Chemistry
1 answer:
Juliette [100K]3 years ago
5 0

Answer:

Explanation:

4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂.

4 moles of FeS₂ reacts with 11 moles of O₂

8 moles of FeS₂ reacts with 22  moles of O₂

Moles of O₂ available is 15 .

So O₂ is in short supply .

So O₂ is the limiting reagent.

Option (c) is the right choice.

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The chemical equation is unbalanced and synthesized.

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The complete question is below:

After learning about the law of conservation of mass, Sammy became interested in balancing equations. He knew that the symbol for aluminum was Al and silver tarnish was Ag2S. He also knew that mixing the two chemicals yielded pure silver, or Ag, in an aluminum sulfide solution. Here is the equation showing this reaction:

3 Ag2S + 2 Al → 6 Ag + Al2S3

This equation is (synthesis / unbalanced / replacement / balanced), and it represents a(n) (unbalanced / balanced / synthesized / replaced) chemical reaction.

answer choices:

  • synthesis; balanced

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6 0
1 year ago
You are stranded on survivor island in which there are already 375 other survivors. For every survivor there are two hands, for
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Answer:

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Explanation:

4 0
3 years ago
What is the formula for Tetrachlorine pentaiodide
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Answer:

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Explanation:

4 0
3 years ago
A sample of neon occupies a volume of 752 ml at 25 0
lyudmila [28]
V_{initial} = 752\:mL
T_{initial} = 25.0^0C
converting to Kelvin
TK = TC + 273
TK = 25.0 + 273 → TK = 298.0 → T_{initial} = 298.0\:K
V_{final} = ? (in\:milliliters)
T_{final} = 50.0^0C
TK = TC + 273
TK = 50.0 + 273 → TK = 323.0 → T_{final} = 323.0\:K

By the first Law of Charles and Gay-Lussac, we have: 
\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }

Solving:
\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }
\frac{ 752 }{ 298.0 } = \frac{ V_{f} }{ 323.0 }
Product of extremes equals product of means:
298.0* V_{f} = 752*323.0
298.0 V_{f} = 242896
V_{f} = \frac{242896}{298.0}
\boxed{\boxed{V_{f} \approx 815.08\:mL}}\end{array}}\qquad\quad\checkmark
5 0
4 years ago
A gas has a volume of 62.65 L at STP. At what temperature (in oC) would the volume of the gas be 78.31 L at a pressure of 612.0
masha68 [24]

Answer:

1.788 C DEGREES

Explanation:

STP is 1 atm at  273.15 K

P1V1/T1  = P2V2/T2

(1)(62.65) / (273.15) = (612/760)(78.31)/T2

T2 = 274.93 K     = 1.788 C

5 0
2 years ago
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