Liquid elemental iron is produced in the thermite reaction because enough heat is generated to melt the iron:
2 Al(s) + Fe2O3(s)----> Al2O3(s) + 2 Fe(l)
What mass of aluminum is needed to produce 750 grams of iron?
To calculate the mass of aluminum needed to produce 750 grams of iron, we need to look at the stoichiometry of the reaction.
From the balanced equation: 2 Al(s) + Fe2O3(s) -> Al2O3(s) + 2 Fe(l), we can see that for every 2 moles of aluminum, we produce 2 moles of iron.
First, we need to determine the molar mass of iron (Fe) and aluminum (Al).
The molar mass of iron (Fe) is approximately 55.845 g/mol.
The molar mass of aluminum (Al) is approximately 26.9815 g/mol.
Next, we can set up a molar ratio using the molar masses:
2 moles Al : 2 moles Fe
26.9815 g/1 mole Al : 55.845 g/1 mole Fe
Now we can solve for the mass of aluminum needed.
Using the given information, we have:
750 g Fe x (2 moles Al / 2 moles Fe) x (26.9815 g Al / 1 mole Al) = 750 g Fe x 26.9815 g Al / 55.845 g Fe
Doing the calculations:
750 g Fe x 26.9815 g Al / 55.845 g Fe ≈ 362.115 g Al
Therefore, approximately 362.115 grams of aluminum are needed to produce 750 grams of iron.
To determine the mass of aluminum needed to produce 750 grams of iron, we need to use the balanced equation for the thermite reaction and apply stoichiometry.
The balanced equation for the reaction is:
2 Al(s) + Fe2O3(s) --> Al2O3(s) + 2 Fe(l)
From the equation, we can see that 2 moles of aluminum react to produce 2 moles of iron.
First, we need to calculate the molar mass of iron (Fe). The molar mass of Fe is approximately 55.85 grams/mole.
Next, we can set up a proportion to find the number of moles of iron (Fe) produced from the given mass:
750 g Fe x (1 mol Fe / 55.85 g Fe) = 13.44 mol Fe
Since 2 moles of aluminum are required to produce 2 moles of iron, the number of moles of aluminum needed is also 13.44 mol.
Finally, we can calculate the mass of aluminum needed using the molar mass of aluminum (Al). The molar mass of Al is approximately 26.98 grams/mole.
Mass of aluminum = number of moles of aluminum x molar mass of aluminum
Mass of aluminum = 13.44 mol Al x 26.98 g/mol Al
Mass of aluminum = 362.01 grams
Therefore, to produce 750 grams of iron, approximately 362.01 grams of aluminum are needed.