PLeaseee ITS Timedddd + BrainlIESTT A 6.50 gram piece of aluminum reacts with an excess of oxygen. use the balanced equation below to

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PLeaseee ITS Timedddd + BrainlIESTT
A 6.50 gram piece of aluminum reacts with an excess of oxygen. use the balanced equation below to determine how many grams of aluminum oxide is formed during this reaction.
4 Al + 3 O2 –> 2 Al2O3
a. 662.7 grams of Al2O3
b. 6.1 grams of Al2O3
c. 24.6 grams of Al2O3
d. 12..3 grams of Al2O3

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Delwyn 5 months 2021-09-03T09:54:24+00:00 1 Answers 0 views 0

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    2021-09-03T09:55:36+00:00

    Answer:

    d. 12.3 grams of Al2O3

    Explanation:

    The balanced chemical equation of this chemical reaction is as follows:

    4Al + 3O2 –> 2Al2O3

    Based on the balanced equation, 4 moles of aluminum (limiting reagent) reacts to form 2 moles of aluminum oxide (Al2O3).

    First, we need to convert the mass of aluminum to moles using the formula;

    mole = mass/molar mass

    Molar mass of Al = 27g/mol

    mole = 6.50/27

    = 0.241mol of Al.

    Hence, if 4 moles of aluminum (limiting reagent) reacts to form 2 moles of aluminum oxide (Al2O3).

    Then, 0.241mol of Al will produce (0.241 × 2/4) = 0.241/2 = 0.121mol of Al2O3.

    Convert this mole value to molar mass using mole = mass/molar mass

    Molar mass of Al2O3 = 27(2) + 16(3)

    = 54 + 48

    = 102g/mol

    mass = molar mass × mole

    mass = 102 × 0.121

    mass of Al2O3 = 12.34grams.

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )