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VMariaS [17]
3 years ago
11

Consider the reaction: A (aq) ⇌ B (aq) at 253 K where the initial concentration of A = 1.00 M and the initial concentration of B

= 0.000 M. At equilibrium it is found that the concentration of B = 0.357 M. What is the maximum amount of work that can be done by this system when the concentration of A = 0.867 M? (in kJ)
Chemistry
1 answer:
Ugo [173]3 years ago
4 0

Answer:

The maximum amount of work that can be done by this system is -2.71 kJ/mol

Explanation:

Maximum amount of work denoted change in gibbs free energy (\Delta G) during the reaction.

Equilibrium concentration of B = 0.357 M

So equilibrium concentration of A = (1-0.357) M = 0.643 M

So equilibrium constant at 253 K, K_{eq}= \frac{[B]}{[A]}

[A] and [B] represent equilibrium concentrations

K_{eq}=\frac{0.357}{0.643}=0.555

When concentration of A = 0.867 M then B = (1-0.867) M = 0.133 M

So reaction quotient at this situation, Q=\frac{0.133}{0.867}=0.153

We know,  \Delta G=RTln(\frac{Q}{K_{eq}})

where R is gas constant and T is temperature in kelvin

Here R is 8.314 J/(mol.K), T is 253 K, Q is 0.153 and K_{eq} is 0.555

So, \Delta G=8.314\times 253\times ln(\frac{0.153}{0.555})mol/K

                           = -2710 J/mol

                            = -2.71 kJ/mol

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For the reaction ? Fe+? H2o ⇀↽? Fe3o4+? H2 , a maximum of how many grams of fe3o4 could be formed from 354 g of fe and 839 g of
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The given reaction is:

3Fe + 4H2O → Fe3O4 + 4H2

Given:

Mass of Fe = 354 g

Mass of H2O = 839 g

Calculation:

Step 1 : Find the limiting reagent

Molar mass of Fe = 56 g/mol

Molar mass of H2O = 18 g/mol

# moles of Fe = mass of Fe/molar mass Fe  = 354/56 = 6.321 moles

# moles of H2O = mass of h2O/molar mass of H2O = 839/18 = 46.611 moles

Since moles of Fe is less than H2O;  Fe is the limiting reagent.

Step 2: Calculate moles of Fe3O4 formed

As per reaction stoichiometry:

3 moles of Fe form 1 mole of Fe3O4

Therefore, 6.321 moles of Fe = 6.321 * 1/ 3 = 2.107 moles of Fe3O4

Step 4: calculate the mass of Fe3O4 formed

Molar mass of Fe3O4 = 232 g/mol

# moles = 2.107 moles

Mass of Fe3O4 = moles * molar mass

= 2.107 moles * 232 g/mol = 488.8 g (489 g approx)

 


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Read 2 more answers
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(b)
Luda [366]

The maximum mass of anhydrous zinc chloride that could be obtained from the products of the reaction is 8.18 g

<h3>Stoichiometry </h3>

From the question, we are to determine the maximum mass of anhydrous zinc chloride that could be obtained

From the given balanced chemical equation,

ZnO + 2HCl → ZnCl₂ + H₂O

This means 1 mole of ZnO will completely react with 2 moles of HCl to produce 1 mole of ZnCl₂ and 1 mole of H₂O

From the given information

Number of moles of ZnO = 0.0830 mole

Now, we will calculate the number of moles of HCl that is present

Volume of HCl added = 100 cm³ = 0.1 dm³

Concentration of the HCl = 1.20 mol/dm³

Using the formula,

Number of moles = Concentration × Volume

Number of moles of HCl present = 1.20 × 0.1 = 0.120 mole

Since

1 mole of ZnO will completely react with 2 moles of HCl to produce 1 mole of ZnCl₂

Then,

0.06 mole of ZnO will react with the 0.120 mole of HCl to produce 0.06 mole of ZnCl₂

Therefore, maximum number of moles of anhydrous zinc chloride that could be produced is 0.06 mole

Now, for the maximum mass that could be produced

Using the formula,

Mass = Number of moles × Molar mass

Molar mass of ZnCl₂ = 136.286 g/mol

Then,

Mass = 0.06 × 136.286

Mass = 8.17716 g

Mass ≅ 8.18 g

Hence, the maximum mass of anhydrous zinc chloride that could be obtained from the products of the reaction is 8.18 g

Learn more on Stoichiometry here: brainly.com/question/11910892

3 0
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