Question

what is the percent yield of reaction in which 455g of tungsten (VI) oxide (WO3) reacts with excess hydrogen gas to produce metallic tungsten and 9.60 cm^3 of water.(density of water = 1.00 g/cm^3) ???
WO3 + 3H2 ==> 3H2O + W

mols WO3 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols WO3 to mols H2O.
Now convert mols H2O to grams H2O. This is the theoretical yield (TY). The actual yield (AY) is given in the problem as 9.60 grams (since the density is 1.00 g/mL).
% yield = (AY/TY)*100 = ?

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