keizhacarbajal6374 keizhacarbajal6374
  • 18-05-2018
  • Chemistry
contestada

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

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ChemistryHelper2024
ChemistryHelper2024 ChemistryHelper2024
  • 27-05-2018
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
₂O = 0.964 / 2 = 0.482 

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

% yield = 39.7 x 100 / 90.1 = 44.1
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