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30 April, 21:08

Determine the percent yield for the reaction between 82.4 g of rb and 11.6 g of o2

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  1. 30 April, 21:36
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    4 Rb + O₂ → 2 Rb₂O

    Number of moles Rb = 82.4 g / 85.47 g/mol = 0.964

    Number of moles O ₂ = 11.6 g / 32 g/mol = 0.363

    the ratio between Rb and O ₂ is 4 : 1 so Rb is the limiting reactant

    Moles Rb ₂ O = 0.964 / 2 = 0.482

    Mass Rb ₂ O = 0.482 mol x 186.94 g/mol = 90.1 g

    % yield = 39.7 x 100 / 90.1 = 44.1
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