J.R. S. answered 04/22/24
Ph.D. University Professor with 10+ years Tutoring Experience
SiO2(s) + 6 HF(aq) H2SiF6(aq) + 2 H2O(l) .. balanced equation
Since we are given the amounts of BOTH reactants, the first thing we must do is to determine which reactant may be in limiting supply. One easy way to do this is to simply divide moles of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less will represent the limiting reactant.
Molar mass SiO2 = 60.1 g / mole
Molar mass HF = 20.0 g / mole
Molar mass H2SiF6 = 144.1 g / mole
Find limiting reactant:
For SiO2: 42.0 g x 1 mol / 60.1 g = 0.6988 moles (÷1 -> 0.699)
For HF: 69.0 g x 1 mol / 20.0 g = 3.450 moles (÷6 -> 0.575)
Since 0.575 is less than 0.699, HF is LIMITING and moles of HF will be used to determine theoretical yield
Theoretical yield of H2SiF6:
3.450 mols HF x 1 mol H2SiF6 / 6 mol HF x 144.1 g H2SiF6 / mol = 82.9 g H2SiF6 (theoretical yield)
Percent yield:
% yield = actual yield / theoretical yield (x100%)
% yield = 45.8 g / 82.9 g (x100%) = 55.2% yield
(be sure to check all of the math)