chemistry equation

Question

Determine the percent yield of the reaction below when a scientist used 40.0 grams of SiO2 and 40.0 grams of HF and was able to yield 45.8 grams of H2SiF6?

SiO2 (s) + 6 HF (aq) =  H2SiF6 (aq) + 2 H2O(l)

Answer 1

Whenever you are given the mass or moles of BOTH reactants, you must determine which reactant, if any, is limiting.

SiO₂ (s) + 6HF (aq) =  H₂SiF₆ (aq) + 2H₂O(l)

40.0g........40.0..............?.............................

40.0 g SiO₂ x 1mol SiO₂/60.1 g x 1 mol H₂SiF₆/mol SiO₂ x 144 g/mol H₂SiF₆ = 95.8 g H₂SiF₆

40.0 g HF x 1 mol HF/20.0 g x 1 mol H₂SiF₆ / 6 mol HF x 144 g/mol H₂SiF₆ = 48 g H₂SiF₆

Therefore, HF is in limiting supply and theoretical yield of H₂SiF₆ = 48.0 g

% yield = actual yield/theoretical yield (x100%) = 45.8 g/48.0 g (x100%) = 95.4% yield