Final answer:
The maximum mass of iron that can be obtained from 240 tonnes of iron oxide (Fe2O3) is 168,000 tonnes, after calculating the molar mass and using stoichiometry to find the amount of iron in moles and converting it to mass.
Step-by-step explanation:
The question is related to stoichiometry and the conservation of mass in chemical reactions, specifically looking at the reduction of iron oxide (Fe2O3) to iron (Fe).
The relative atomic masses are given as oxygen (O) = 16 and iron (Fe) = 56.
The first step is to calculate the molar mass of iron oxide which is (2 × 56) + (3 × 16) = 160 g/mol.
Then, determine how many moles of iron oxide are present in 240 tonnes (or 240,000,000 grams) by dividing the mass of iron oxide by its molar mass:
240,000,000 g ÷ 160 g/mol = 1,500,000 mol of Fe2O3
Using the stoichiometric ratios from the chemical formula, we know that each mole of Fe2O3 contains 2 moles of Fe. Therefore, the total moles of Fe is:
1,500,000 mol Fe2O3 × 2 mol Fe/mol Fe2O3 = 3,000,000 mol Fe
Finally, convert the moles of Fe to mass:
3,000,000 mol Fe × 56 g/mol Fe = 168,000,000 g or 168,000 tonnes of Fe
This represents the maximum mass of iron that can be obtained from 240 tonnes of iron oxide, assuming complete reduction, with no losses or side reactions.