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Andrei [34K]
2 years ago
7

A student wants to prove the concept of limiting reactants to his lab group during class. If the student adds 56 g or Fe to 71 g

of Cl2, what will the likely result of the reaction be in regards to limiting/excess reactants?
Chemistry
1 answer:
Alexeev081 [22]2 years ago
7 0

The result of the reaction will be such that Cl2 is limiting in its availability

<h3>Limiting reactants</h3>

From the equation of the reaction:

2Fe + 3Cl2 ---> 2FeCl3

Mole ratio of Fe and Cl2 = 2:3

Mole of 56 g of Fe = 56/56 = 1

Mole of 71 g of Cl2 = 71/71 = 1

Whereas, 1 mole of Fe requires 1.5 moles of Cl2. Thus, Cl2 is limiting in availability.

More on limiting reactants can be found here: brainly.com/question/14225536

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A silver cube with an edge length of 2.38 cm and a gold cube with an edge length of 2.79 cm are both heated to 81.9 ∘C and place
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Answer:

The final temperature of the water when thermal equilibrium is reached is 23.54 ⁰C

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<u>Given data;</u>

edge length of silver, a = 2.38 cm = 0.0238 m

edge length of gold, a = 2.79 cm = 0.0279 m

final temperature of silver, t = 81.9 ° C

final temperature of Gold, t = 81.9 ° C

initial temperature of water, t = 19.6 ° C

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<u>Known data:</u>

density gold 19300 kg/m³

density silver 10490 kg/m³

density water 1000 kg/m³

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specific heat silver is 240 J/kgC

specific heat water is 4200 J/kgC

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Apply Pythagoras theorem to determine the side of each cube;

Silver cube;

let L be the side of the silver cube

Taking the cross section of the cube (form a right angled triangle), the edge  length forms the <em>hypotenuse side</em>.

L² + L² = 0.0238²

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Volume of the silver cube = (0.0168)³ = 4.742 x 10⁻⁶ m³

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2L² = 0.0279²

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Mass of silver cube;

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mass = density x volume

mass of silver cube = 10490 (kg/m³) x 4.742 x 10⁻⁶ (m³) = 0.0497 kg

Mass of Gold cube

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Mass of water

mass of water = 1000 (kg/m³) x 0.0001095 (m³) = 0.1095 kg

Let the heat gained by cold water be  Q₁

Let the heat lost  by silver cube = Q₂

Let the heat lost  by gold cube = Q₃

Let the final temperature of water = T

Q₁  = 0.1095 kg x 4200 J/kgC x (T–19.6)

Q₂ = 0.0497 kg x 240 J/kgC x (81.9–T)

Q₃ = 0.148 kg x 129 J/kgC x (81.9–T)

At thermal equilibrium;

Q₁ = Q₂ + Q₃

0.1095 x 4200  (T–19.6)  = 0.0497 x 240 x (81.9–T)  + 0.148 x 129 x (81.9–T)

459.9 (T–19.6) = 11.928 (81.9–T) + 19.092(81.9–T)

459.9T - 9014.04 = 976.9032 - 11.928T + 1563.6348 - 19.092T

459.9T + 11.928T + 19.092T = 9014.04  + 976.9032 + 1563.6348

490.92T = 11554.578

T = 11554.578 / 490.92

T = 23.54 ⁰C

Therefore, the final temperature of the water when thermal equilibrium is reached is 23.54 ⁰C

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