Final answer:
In the thermite reaction between 10.0g of iron(III) oxide and 10.0g of aluminum, the limiting reactant is iron(III) oxide, resulting in the production of 6.99 grams of molten iron.
Step-by-step explanation:
Determining the Limiting Reactant and Mass of Iron Produced in a Thermite Reaction
When 10.0g of iron(III) oxide (Fe2O3)reacts with 10.0g of aluminum (Al), a thermite reaction occurs. This exothermic reaction is represented by the balanced chemical equation:
2 Al(s) + Fe2O3(s) → 2 Fe(s) + Al2O3(s)
To find out the limiting reactant, we need to calculate the moles of each reactant from their given masses. The molar mass of Al is 26.98 g/mol, and the molar mass of Fe2O3 is 159.69 g/mol.
10.0 g Al x (1 mol Al / 26.98 g Al) = 0.3707 mol Al
10.0 g Fe2O3 x (1 mol Fe2O3 / 159.69 g Fe2O3) = 0.0626 mol Fe2O3
According to the balanced chemical equation, the stoichiometry is 2 moles of Al per 1 mole of Fe2O3. Dividing each number of moles by the respective stoichiometric coefficient:
0.3707 mol Al / 2 = 0.1854
0.0626 mol Fe2O3/ 1 = 0.0626
The smaller result indicates the limiting reactant. Hence, Fe2O3 limits the reaction and determines the maximum amount of product.
As per the equation, 1 mole of Fe2O3 yields 2 moles of Fe. Therefore, 0.0626 mol Fe2O3 yields 2 x 0.0626 mol of Fe = 0.1252 mol Fe. The molar mass of Fe is 55.85 g/mol.
0.1252 mol Fe x (55.85 g Fe / 1 mol Fe) = 6.99 g Fe (rounded to three significant figures)
So, 6.99 grams of molten iron is produced from the thermite reaction between 10.0g of iron(iii) oxide and 10.0g of Al, with Fe2O3 being the limiting reactant.