15. 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?
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?
2 moles of Al produces 2 moles of Fe. and 750grams of Al is 750g/26gmol-1 = 28.85moles. 2Al is 2x28.85moles = 57.7moles which is equal to the mole of Fe.
with this mole, use the formula m = nMr to find the mass, where n is the mole and Mr is the molar mass of iron.
To find the mass of aluminum needed to produce 750 grams of iron, we need to use the balanced chemical equation and convert grams of iron into moles, and then use the molar ratio to determine the moles of aluminum needed. Finally, we can convert the moles of aluminum into grams.
Step 1: Write out the balanced chemical equation
2 Al(s) + Fe2O3(s) → Al2O3(s) + 2 Fe(l)
Step 2: Convert grams of iron into moles
We know from the balanced equation that 2 moles of aluminum will produce 2 moles of iron. Therefore, the molar mass of aluminum is equal to the molar mass of iron.
Molar mass of iron (Fe) = 55.845 g/mol
To find the moles of iron, we divide the given mass of iron by the molar mass of iron:
Moles of iron = Mass of iron (g) / Molar mass of iron (g/mol)
Moles of iron = 750 g / 55.845 g/mol
Step 3: Determine the moles of aluminum needed
Since the molar ratio between aluminum and iron is 1:1, the moles of aluminum needed will be the same as the moles of iron. Therefore, the moles of aluminum needed is equal to the moles of iron.
Moles of aluminum needed = Moles of iron = 750 g / 55.845 g/mol
Step 4: Convert moles of aluminum into grams
To convert the moles of aluminum into grams, we need to use the molar mass of aluminum:
Molar mass of aluminum (Al) = 26.98154 g/mol
Now we can calculate the mass of aluminum needed:
Mass of aluminum needed = Moles of aluminum needed x Molar mass of aluminum
Mass of aluminum needed = (750 g / 55.845 g/mol) x 26.98154 g/mol
Finally, calculate the mass of aluminum needed by substituting the values into the equation:
Mass of aluminum needed = 12.74 g
Therefore, approximately 12.74 grams of aluminum is needed to produce 750 grams of iron.