Question

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

Answer 1

4 Rb + O₂ → 2 Rb₂O

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

Number of moles O₂ = 11.6 g / 32 g/mol =0.363

the ratio between Rb and O₂ is 4 : 1 so Rb is the limiting reactant

Moles Rb₂O = 0.964 / 2 = 0.482

Mass Rb₂O = 0.482 mol x 186.94 g/mol = 90.1 g

% yield = 39.7 x 100 / 90.1 = 44.1