Question

The thermite reaction, used for welding iron, is the reaction of Fe3O4 with Al. 8 Al (s) + 3 Fe3O4 (s) ⟶ 4 Al2O3 (s) + 9 Fe (s) Δ H° = -3350. kJ/mol rxn. Because this large amount of heat cannot be rapidly dissipated to the surroundings, the reacting mass may reach temperatures near 3000. °C. How much heat (in kJ) is released by the reaction of 19.3 g of Al with 63.2 g of Fe3O4?

Answer 1

The amount of heat released by the reaction is -2395.25 kJ.

Number of moles:

It is defined as the ratio of given mass over molar mass.

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

For aluminum:

Given mass= 19.3 g

Molar mass = 27 g/mol

→ Calculation for number of moles:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\\\\\text{Number of moles}=\frac{19.3g}{27g/mol} \\\\\text{Number of moles}=0.715 mol$

For Ferric oxide  :

Given mass = 63.2 g

Molar mass = 231.5 g/mol

→ Calculation for number of moles:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\\\\\text{Number of moles}=\frac{63.2g}{231.5g/mol} \\\\\text{Number of moles}=0.273 mol$

Chemical equation:

$8Al (s) + 3 Fe_3O_4 (s)----> 4 Al_2O_3 (s) + 9 Fe (s)$

According to the reaction:

8 moles of aluminum reacts with 3 moles of ferric oxide

So, 0.715 moles of aluminum will react with = $\frac{3}{8}*0.715 =0.268mol$ of ferric oxide

As, given amount of ferric oxide is more than the required amount.

So, it is considered as an excess reagent. And, aluminum is considered

→ Calculation of enthalpy change:

$\triangle H_{rxn}=\frac{q}{n}$

where,

q = amount of heat absorbed = ?

n = number of moles of aluminum = 0.715 moles

$\triangle H_{rxn}$ = enthalpy change of the reaction  = -3350 kJ/mol

Substituting the values in the above formula:

$\triangle H_{rxn}=\frac{q}{n}\\\\q= \triangle H_{rxn}*n\\\\q=-3350kJ/mol*0.715mol\\\\q=-2395.25kJ$

Hence, the amount of heat released by the reaction is -2395.25 kJ.

Find more information about number of moles here:

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