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how many grams of Al2O3 be obtained if 13.5 g of aluminum completely reacts with oxygen as

4Al+3O2---->2Al2O3
Molar mass of Al is 27g/mol.

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4 answers

  1. n=given mass/reaction mass=13.6/27=1/2
    Now.,
    n=given mole/reaction mole(Al2O3)=1/2/2=1/4
    Now.,
    n=mass/molar mass(Al2O3)
    mass=1/4*102=25.50grams

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  2. 4Al +3O2-------------> 2Al2O3
    4Al----------->2Al2O3
    1Al---------> 2/4 Al2O4
    First convert grams to moles
    Moles of aluminum = 13.5/27 =0.5
    Now we have moles
    0.5 moles of Al = 0.5 ×2/ 4 = p

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  3. 4Al +3O2-------------> 2Al2O3
    4Al----------->2Al2O3
    1Al---------> 2/4 Al2O4
    First convert grams to moles
    Moles of aluminum = 13.5/27 =0.5
    Now we have moles
    0.5 moles of Al = 0.5 ×2/ 4 = 1/4
    So we have moles of Al2O3 now but in question they asked us in grams. So we have to convert it to grams
    Grams = moles x molar mass of Al2O3
    Grams = 1/4 × 102 = 25.5

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  4. 4Al+3O2---->2Al2O3

    mols Al = grams Al/atomic mass Al.
    Using the coefficients in the balanced equation, convert mols Al to mols Al2O3. You can see that 2 mol Al2O3 are formed for every 4 mols Al initially.
    Then convert mols Al2O3 to grams. g = mols x molar mass = ?
    Post your wor if you get stuck

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