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11 December, 08:06

How many grams of Boron can be obtained from 234 grams of B2O3?

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  1. 11 December, 10:22
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    72.67g of B

    Explanation:

    The reaction of B₂O₃ to produce boron (B), is:

    B₂O₃ → 3/2O₂ + 2B

    That means B₂O₃ produce 2 moles of boron

    Molar mass of B₂O₃ is 69.62g/mol. 234g of B₂O₃ contains:

    234g B₂O₃ ₓ (1mol / 69.62g) = 3.361 moles of B₂O₃.

    As 1 mole of B₂O₃ produce 2 moles of B, Moles of B that can be produced from B₂O₃ is:

    3.361mol B₂O₃ ₓ 2 = 6.722 moles of B.

    As molar mass of B is 10.811g/mol. Thus mass of B that can be produced is:

    6.722mol B ₓ (10.811g / mol) = 72.67g of B
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