Given the Following reaction Fe(s) + O2(g) --> Fe2O3(s)?

a. How many moles of Fe2O3 can be formed if 2.0 moles of Fe react?

b. How many grams of O2 are needed to react with 2.0 moles of Fe?

c. How many grams of Fe2O3 can be formed if 5.0g of Fe react?

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  • 1 decade ago
    Favorite Answer

    4Fe(s) + 3O2(g) --> 2Fe2O3(s)

    2.0 mol Fe * 2 mol Fe2O3 / 4 mol Fe = 1 mol Fe2O3

    (find the mole ratios from the coefficients of the chemical equation)

    2.0 mol Fe * 3 mol O2 / 4 mol Fe = 1.5 mol O2

    5.0 g Fe * 1 mol Fe / 55.85 g Fe * 2 mol Fe2O3 / 4 mol Fe * 159.7 g Fe2O3 / 1 mol Fe2O3

    = 7.15 g

    Steps for mass-mass conversions:

    mass of known -> moles of known (via the molar mass of known) -> find molar ratio between known and unknown -> mass of unknown (via the molar mass of unknown)

    mole-mole:

    moles of known-> molar ratio between known and unknown-> moles of unknown (answer)

    :)

  • Dr.A
    Lv 7
    1 decade ago

    4 Fe +3 O2 >> 2Fe2O3 is the balanced equation

    a. the ratio between Fe and Fe2O3 is 4 : 2. 1.0 mole of Fe2O3 can be formed if 2.0 moles of Fe react

    b. the ratio between Fe and O2 is 4 : 3

    moles O2 needed = 2.0 x 3 / 4 = 1.5

    mass O2 = 1.5 mol x 32 g/mol = 48.0 g

    c. Moles Fe = 5.0 g / 55.847 g/mol = 0.090 ( 2 sign. fig.)

    The ratio between Fe and Fe2O3 is 4 : 2

    Moles Fe2O3 = 0.090 x 2 / 4 = 0.045

    Mass Fe2O3 = 0.045 mol x 159.694 g/mol = 7.2 g

  • Anonymous
    1 decade ago

    4fe + 3o2 ---> 2fe2o3. - balanced..

    ratio of fe and fe2o3 = 4:2

    so, i mol. can be formed if two moles react.........

    a). 1 mol..

    ratio bet. fe and o2 = 4:3

    moles. = 2.0 x 3/4 =1.5

    1.5 x 32g/mol. = 48....

    since o2 = 16 x 16 = 32 x 105 = 48g

    b).48g.

    ratio - fe- fe2o3 = 4:2.

    moles fe2 o3 .= 0.90 x 2/4 = 0.045..

    sice fe = 5 /5.847 = 0.0895 = 0.090.

    mass = 0.045 x 159.69 of g/mol.

    = 7.18605 = 7.2....

    i m studying 8th

  • 1 decade ago

    Balance the equation: 4Fe + 3O2 -> 2Fe2O3

    MW of Fe = 55.845 g/mol

    MW of O2 = 31.999 g/mol

    MW of Fe2O3 = 159.69 g/mol

    a) According to the balanced equation, there is 4 mol of Fe for 2 mol of Fe2O3. Therefore:

    (2.0 mol Fe)(2 mol Fe2O3 / 4 mol Fe) = 1.0 mol Fe2O3

    b)

    (2.0 mol Fe)(3 mol O2 / 4 mol Fe) = 1.5 mol O2

    (1.5 mol O2)(31.999 g/mol) = 48.0 g O2

    c)

    (5.0 g Fe) / (55.845 g/mol) = 0.089534 mol Fe

    (0.089534 mol Fe)(2 mol Fe2O3 / 4 mol Fe) = 0.044767 mol Fe2O3

    (0.044767 mol Fe2O3)(159.69 g/mol) = 7.1488 g Fe2O3

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    5 years ago

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