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3 years ago

Given the following equation: 8 Fe + S8 —> 8 FeS what mass of iron is needed to react with 5.65 mol of sulfur S8

Chemistry

1 answer:

Mekhanik [1.2K]3 years ago

4 0

Answer:

2522g

Explanation:

8Fe + S8 —> 8FeS

From the question,

8moles of Fe required 1mole of S8.

Therefore, Xmol of Fe will require 5.65mol of S8 i.e

Xmol of Fe = 8 x 5.65 = 45.2moles

Now we need to covert 45.2moles to gram. This is illustrated below below:

Molar Mass of Fe = 55.8g/mol

Number of mole of Fe = 45.2moles

Mass of Fe =?

Mass = number of mole x molar Mass

Mass of Fe = 45.2 x 55.8

Mass of Fe = 2522g

Therefore, 2522g of Fe is needed for the reaction.

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1.04 ⨯ 10 $^{-4}$ M

<h3>Explanation</h3>

A = ε $\cdot l \cdot c$ by the Beer-Lambert law, where

A the absorbance,
$l$ the path length,
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$c$ concentration of the solution.

A and ε are the same for both solutions. Therefore, $l \cdot c$ is constant; $l$ is inversely proportional to $c$ . The 100 mL sample would have a concentration 1/4.78 times that of the 45.0 mL reference.

The 13.0 mL standard solution has a concentration of 5.17 ⨯ $10^{-4}$ M. Diluting it to 45.0 mL results in a concentration of $5.17 \times 10^{-4} \times \frac{13.0}{45.0} =$ 1.494 M.

$c$ is inversely related to $l$ for the two solutions. As a result, c₂ = $c_1 \cdot \frac{l_1}{l_2} = 1.494 \times 10^{-4} \times \frac{1}{4.78} =$ 3.126 M.

The 30.0 mL sample has to be diluted by 30.0 / 100.0 times to produce the 100.0 mL solution being tested. The 100.0 mL solution has a concentration of 3.126 M. Therefore, the 30.0 mL solution has a concentration of $3.126 \times \frac{100.0}{30.0} =$ 1.04 ⨯ $10^{-4}$ M.

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3 years ago

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Answer:

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3 years ago

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Answer:

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Explanation:

See the graph for this question on the figure attached.

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