Question

Calculate the amount of heat released in the combustion of 10.5 grams of Al with 3 grams of O2 to form Al2O3(s) at 25°C and 1 atm. ΔHfAl2O3(s) = ­−1676 kJ/mol HINT: What does ΔHfAl2O3(s) mean?

Answer 1

Answer : The amount of heat released in the combustion is, 209.5 kJ

Explanation :

First we have to calculate the moles of Al and $O_2$.

$\text{ Moles of }Al=\frac{\text{ Mass of }Al}{\text{ Molar mass of }Al}=\frac{10.5g}{27g/mole}=0.389moles$

$\text{ Moles of }O_2=\frac{\text{ Mass of }O_2}{\text{ Molar mass of }O_2}=\frac{3g}{32g/mole}=0.188moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

$4Al+3O_2\rightarrow 2Al_2O_3$

From the balanced reaction we conclude that

As, 3 mole of $O_2$ react with 4 mole of $Al$

So, 0.188 moles of $O_2$ react with $\frac{4}{3}\times 0.188=0.251$ moles of $Al$

From this we conclude that, $Al$ is an excess reagent because the given moles are greater than the required moles and $O_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Al_2O_3$

From the reaction, we conclude that

As, 3 mole of $O_2$ react to give 2 mole of $Al_2O_3$

So, 0.188 moles of $O_2$ react to give $\frac{2}{3}\times 0.188=0.125$ moles of $Al_2O_3$

Now we have to calculate the amount of heat released in the combustion.

As, 1 mole of $Al_2O_3$ releases amount of heat = 1676 kJ

So, 0.125 mole of $Al_2O_3$ releases amount of heat = $0.125\times 1676kJ=209.5kJ$

Thus, the amount of heat released in the combustion is, 209.5 kJ

Answer 2

The amount of heat released in the combustion is 209.5 kJ.

Number of moles:

It is defined as given mass over molar mass.

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For Aluminum:

Given mass = 10.5 g

Molar mass = 27g/mol

→  Calculation for number of moles:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\\\\\text{Number of moles}=\frac{10.5g}{27g/mol}\\\\ \text{Number of moles}=0.389moles$

For Oxygen:

Given mass = 3 g

Molar mass = 16 g/mol

→  Calculation for number of moles:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\\\\\text{Number of moles}=\frac{3g}{16g/mol}\\\\ \text{Number of moles}=0.188moles$

Balanced chemical reaction:

$4Al+3O_2--->Al_2O_3$

→ According to the reaction:

3 mole of oxygen react with 4 mole of aluminum

So, 0.188 moles of oxygen react with $\frac{4}{3}*0.188=0.251$ moles of aluminum

Thus, we can conclude that aluminum is an excess reagent because the given moles are more than the required moles and Oxygen is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Al_2O_3$ :

→ According to the reaction:

3 mole of oxygen react to give 2 mole of $Al_2O_3$

So, 0.188 moles of  react to give $\frac{2}{3}*0.188=0.125$ moles of  $Al_2O_3$ .

Now we have to calculate the amount of heat released in the combustion.

As, 1 mole of  $Al_2O_3$ releases amount of heat = 1676 kJ

So, 0.125 mole of  $Al_2O_3$   releases amount of heat = 0.128*1676=209.5kJ

Thus, the amount of in the combustion is 209.5 kJ.

Find more information about Combustion here:

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