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a) Initial mass of irong oxide: 6,310gb) Initial mass of Aluminum: 2,170g

Explanation:

It is necessary to use the percent yield of the experiment. Since it is not provided, I will use a hypothetical percent yield of 90%.

1. Write the chemical equation:

     

2. Write the mole ratios:

     

3. Covert 5.00 kg of soid iron to number of moles (n)

n = mass in grams / atomic massatomic mass of Fe = 55.845g/mol(5.00kg × 1,000g/kg) ÷ (55.845 g/mol) = 89.5335mol Fe

4. Use the mole ratios to find the number of moles of each reactant

     

     

5.  Use the molar masses to convert the number of moles into mass:

mass = number of moles × molar mass

Fe₂O₃:

molar mass: 159.69 g/molmass = 44.76675mol × 156.69g/mol = 7,014.50g

Al:

atomic mass: 26.982g/molmass = 89.5335mol × 26.982g/mol = 2,415.79g

6. Multiply by the percent yield: 90% (assumed)

Fe₂O₃:

7,014.50g × 90% = 6,313.05g = 6,310g (three significant figures)

Al:

2,415.79g × 90% = 2,174.21g = 2,170g (three significant figures)

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Answer ozone (o3) is made of three atoms of oxygen, it is a molecule;