chemistry

can you calculate the mass(g) of AL2O3 that is produced when 2.5g of aluminum and 2.5g of oxygen react according to this equation : 4Al(s)+3O2(g)->2Al2O3(s)

im not understanding how to do this

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  1. Yes, I can.

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  2. could you help me

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  3. @DrBob222

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  4. This procedure will work almost all of the stoichiometry problems. This one is a limiting reagent problem (LR); you know that because amounts are given for BOTH rectants. I work these the long way.

    1. Write and balance the equation. You have that. 4Al(s)+3O2(g)->2Al2O3(s)

    2. Convert what you have into mols.
    a. mols Al = grams/molar mass = ?
    b. mols O2 = grams/molar mass = ?

    3. Use the coefficients in the balanced equation to convert mols Al and mols O2 to mols Al2O3.
    a. mols Al to mols Al2O3 is mols Al x (2 mol Al2O3/4 mols Al) = ?
    b. Do the same for O2.
    c. You see the mols Al2O3 are not equal from the two calculations which means one of them is not right; the correct value in LR problems is ALWAYS the smaller number and the reagent producing that number is the LR.

    4. Now convert mols to grams using the smaller LR reagent.
    g = mols x molar mass = ?

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  5. how do i convert into mols?

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  6. 2a and 2b above. mols = grams/molar mass
    For Al that is grams/atomic mass = 2.5 g/ 27 = ?

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  7. i got 0.09259

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  8. I don't get that answer. Post your work and I'll find the error. Follow the steps above.

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  9. Your 0.09259 is mols Al. That's the first step. Now you do mols O2 and follows steps 2,3,4.

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  10. for 02 i got 0.078125
    im not really under standing what you want me to do for step 2? didn't i ustconvert them into mols?

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  11. You must read and follow step by step. I'll get you started. Following are my step by step instructions and I'll do the first few steps to show you how it's done.

    1. Write and balance the equation. You have that. 4Al(s)+3O2(g)->2Al2O3(s)

    2. Convert what you have into mols.
    a. mols Al = grams/molar mass = ?
    b. mols O2 = grams/molar mass = ?

    2a. mols Al = grams/atomic mass = 2.5/27 = .0926 mols Al.
    2b. mols O2 = grams/molar mass = 2.5/32 = 0.0781 mols O2


    3. Use the coefficients in the balanced equation to convert mols Al and mols O2 to mols Al2O3.
    a. mols Al to mols Al2O3 is mols Al x (2 mol Al2O3/4 mols Al) = ?
    0.0926 mols Al x (2 mols Al2O3/4 mols Al) = 0.0926 x (2/4) = 0.0463 mols Al2O3 if we had 2.5 g Al and all of the O2 we needed.

    b. Do the same for O2.
    0.0781 mols O2 x (2 mols Al2O3/3 mols O2) = 0.0781 x (2/3) = 0.0521 mols Al2O3 if we had 2.5 g O2 and all of the Al we needed.
    c. You see the mols Al2O3 are not equal from the two calculations which means one of them is not right; the correct value in LR problems is ALWAYS the smaller number and the reagent producing that number is the LR.
    Now you do step 3c and 4. Post your work if you get stuck.

    4. Now convert mols to grams using the smaller LR reagent.
    g = mols x molar mass = ?

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  12. OK were going to use 0.0521 because its smaller than 0.0463. so 0.0521/32?

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  13. wait sorry i got it thank you for being so patient with me

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  14. the answer is mass Al2O3 = 101.96 g/mol x 0.04633 moles
    = 4.72 g

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  15. Yea! Go to the top of the class.

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