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Mars2501 [29]
2 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]2 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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3.5g of a Certain compound X, known to be made of carbon, hydrogen, and perhaps oxygen, and to have a molecular molar mass of 15
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Answer:

C₅H₁₀O₅

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\text{Mass of C} = \text{5.13 g CO}_{2}\times \dfrac{\text{12.01 g C}}{\text{44.01 g }\text{CO}_{2}}= \text{1.400 g C}

(b) Mass of H

\text{Mass of H} = \text{2.10 g H$_{2}$O}\times \dfrac{\text{2.016 g H}}{\text{18.02 g H$_{2}$O}} = \text{0.2349 g H}

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Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g

2. Calculate the moles of each element

\text{Moles of C = 1400  mg C}\times\dfrac{\text{1 mmol C}}{\text{12.01 mg C }} = \text{116.6 mmol C}\\\\\text{Moles of H = 234.9 mg H} \times \dfrac{\text{1 mmol H}}{\text{1.008 mg H}} = \text{233.1 mmol H}\\\\\text{Moles of O = 1870 mg O} \times \dfrac{\text{1 mmol O}}{\text{16.00 mg O}} = \text{116 mmol O}

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

\text{C: } \dfrac{116.6}{116.6}= 1\\\\\text{H: } \dfrac{233.1}{116.6} = 1.999\\\\\text{O: } \dfrac{116}{116.6} = 1.00

4. Round the ratios to the nearest integer

C:H:O = 1:2:1

5. Write the empirical formula

The empirical formula is CH₂O.

6. Calculate the molecular formula.

EF Mass = (12.01 + 2.016  + 16.00) u  = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{150 u}}{\text{30.03 u}} = 5.00  \approx 5

MF = (CH₂O)₅ = C₅H₁₀O₅

The molecular formula of X is C₅H₁₀O₅.

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