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Ipatiy [6.2K]
3 years ago
10

A 6.55 g sample of aniline (C6H5NH2, molar mass = 93.13 g/mol) was combusted in a bomb calorimeter with a heat capacity of 14.25

kJ/°C. If the initial temperature was 32.9°C, use the information below to determine the value of the final temperature of the calorimeter.
4 C6H5NH2(l) + 35 O2(g) → 24 CO2(g) + 14 H2O(g) + 4 NO2(g) ΔH°rxn= -1.28 x 104 kJ
Chemistry
1 answer:
Ganezh [65]3 years ago
3 0

Answer:

Final temperature = 48.6867 °C

Explanation:

The expression for the heat of combustion in bomb calorimeter is:

<u>ΔH = -C ΔT </u>

where,

ΔH is the enthalpy of the reaction

C is the heat capacity of the bomb calorimeter

ΔT is the temperature change

Given in the question:

ΔH°rxn = -1.28 x 10⁴ kJ

From the balanced reaction, 4 moles of aniline reacting with oxygen. Thus, enthalpy change of the reaction in kJ/mol is:

<u>ΔH = -12800 kJ/4 = -3200kJ/mole </u>

Given:

Mass of aniline combusted = 6.55 g

Molar mass of aniline = 93.13 g/mol

Thus moles of aniline = 6.55 / 93.13 moles = 0.0703 moles

The total heat released from 0.0703 moles of aniline is

<u>ΔH = -3200kJ/mole x 0.0703 moles = -224.96 kJ </u>

Given: Heat capacity of calorimeter is 14.25 kJ/°C

T₁ (initial) = 32.9°C

T₂ (final) = ?

From the above formula:

-224.96 kJ = -14.25kJ/°C (T₂ - 32.9)

Solving for T₂ , we get:

<u>T₂ = 48.6867 °C</u>

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<u>Answer:</u> The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       ....(1)

  • <u>For Iron:</u>

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:  

\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol

For the given chemical reaction:

2Fe(s)+O_2(g)\rightarrow 2FeO(s)

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = \frac{2}{2}\times 0.0771=0.0771mol of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:  

0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g

To calculate the percentage yield of iron (ii) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

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