The thermite reaction is a very exothermic reaction; it has been used to produce liquid iron for welding. A mixture of 2 mol of powdered aluminum metal and 1 mol of iron(III) oxide yields liquid iron and solid aluminum oxide. How many grams of the mixture are needed to produce 313 kJ of heat?
2Al + Fe2O3 ==>2Fe and Al2O3 + ??kJ heat.
Do you know the kJ heat produced? If so,
313 kJ/?? kJ = moles Fe2O3 needed and convert that to grams Fe2O3.
(313 kJ/?? kJ) x 2 = moles Al needed and convert that to grams Al.
Add the grams Fe2O3 and grams Al to get the grams of the mixture needed.
To find the number of grams of the mixture needed to produce 313 kJ of heat, we need to use the stoichiometry of the thermite reaction to determine the amount of reactants required.
First, let's write out the balanced chemical equation for the thermite reaction:
2Al + Fe2O3 -> Al2O3 + 2Fe
From the equation, we can see that 2 moles of aluminum react with 1 mole of iron(III) oxide to produce 2 moles of iron and 1 mole of aluminum oxide.
Next, we need to calculate the molar enthalpy change (ΔH) for the reaction. Given that it is an exothermic reaction, the ΔH value will be negative.
The ΔH value for the thermite reaction is -851.5 kJ/mol of iron(III) oxide.
Now, we can use the stoichiometry and ΔH value to calculate the number of moles of iron(III) oxide required to produce 313 kJ of heat:
313 kJ * (1 mol Fe2O3 / -851.5 kJ) = -0.3675 mol Fe2O3
Since the stoichiometric ratio between aluminum and iron(III) oxide is 2:1, we need twice the number of moles of aluminum. Therefore,
2 * -0.3675 mol Fe2O3 = -0.735 mol Al
To convert the moles of aluminum into grams, we need to know the molar mass of aluminum. The molar mass of aluminum is approximately 26.98 g/mol.
-0.735 mol Al * 26.98 g/mol = -19.5 g Al
Since we cannot have a negative mass, we can conclude that approximately 19.5 grams of the mixture (consisting of 2 mol of aluminum and 1 mol of iron(III) oxide) are needed to produce 313 kJ of heat.
To determine the amount of mixture needed to produce 313 kJ of heat, we first need to calculate the enthalpy change (ΔH) of the reaction.
The balanced equation for the thermite reaction between aluminum and iron(III) oxide is:
2Al + Fe2O3 -> Al2O3 + 2Fe
The enthalpy change for this reaction is -851.5 kJ/mol of Fe2O3.
To calculate the number of moles of Fe2O3 needed to produce 313 kJ of heat, we use the following formula:
moles of Fe2O3 = (kJ of heat) / (enthalpy change per mole)
moles of Fe2O3 = 313 kJ / (-851.5 kJ/mol)
moles of Fe2O3 = -0.367 mol
Since the stoichiometry of the reaction is 1 mol Fe2O3 to 2 mol Al, we need twice as many moles of Al as Fe2O3.
moles of Al = 2 * moles of Fe2O3
moles of Al = 2 * (-0.367 mol)
moles of Al = -0.734 mol
However, moles cannot be negative, so the negative sign is disregarded.
Thus, we need 0.734 moles of aluminum and 0.367 moles of iron(III) oxide to produce 313 kJ of heat.
Next, we can calculate the mass of the mixture using the molar masses of aluminum and iron(III) oxide.
The molar mass of aluminum (Al) is 26.98 g/mol.
The molar mass of iron(III) oxide (Fe2O3) is 159.69 g/mol.
mass of aluminum = moles of Al * molar mass of Al
mass of aluminum = 0.734 mol * 26.98 g/mol
mass of aluminum = 19.78 g
mass of iron(III) oxide = moles of Fe2O3 * molar mass of Fe2O3
mass of iron(III) oxide = 0.367 mol * 159.69 g/mol
mass of iron(III) oxide = 58.59 g
Finally, the total mass of the mixture needed to produce 313 kJ of heat is the sum of the masses of aluminum and iron(III) oxide:
mass of mixture = mass of aluminum + mass of iron(III) oxide
mass of mixture = 19.78 g + 58.59 g
mass of mixture = 78.37 g
Therefore, 78.37 grams of the mixture are needed to produce 313 kJ of heat.