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2 November, 22:56

When oxygen (O) reacts with iron (Fe), it produces rust (Fe2O3). In a closed system, how much oxygen must react with 120 grams of iron to produce 580 grams of rust?

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  1. Yesterday, 00:40
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    It is not possible to produce 580 grams of rust with 120 grams of iron.

    Explanation:

    Here, I show you why it is not possible to produce 580 grams of rust with 120 grams of iron.

    A. Molar masses of the substances:

    Fe: 55.845g/mol 2.1488 mol O₂: 15.999 g/mol Fe₂O₃: 159,69 g/mol 3.6320 mol

    B. The chemical equation is:

    4Fe + 3O₂ → 2Fe₂O₃

    How many grams of iron are there in 580 grams of rust (Fe₂O₃) ?

    1. Convert the mass of Fe₂O₃ into number of moles:

    moles = mass in grams / molar mass moles = 580g / (159.69 g/mol) = 3.6320 mol of Fe₂O₃

    2. Calculate the number of moles of Fe in 3.6320 mol of Fe₂O₃

    There are 2 moles of Fe per mole of Fe₂O₃:

    2 mol Fe / mol Fe₂O₃ * 3.6320 mol Fe₂O₃ = 7.2640 mol Fe

    3. Calculate the mass of 7.2640 moles of Fe:

    mass = number of moles * atomic mass mass = 7.2640 mol * 55.845 g/mol = 405.66 g

    Thus, 120 grams of Fe is too little to produce the mentioned amount of rust; you need about 406 grams of Fe to produce 580 grams of rust.
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