Asked by Jake
What is the % yield of of a reaction in which 40.00g of tungsten(VI)oxide (WO3) reacts with excess hydrogen gas to produce metalic tungsten and 5.40mL of water. (assume the density of water is 1.00g/mL)
Answers
Answered by
DrBob222
WO3 + 3H2 ==> W + 3H2O
mols WO3 = grams/molar mass = approx 0.17
Using the coefficients in the balanced equation, convert mols WO3 to mols H2O. That's ?mols WO3 x (3 mols H2O/1 mol WO3) = ?mols WO3 x 3 = xx mols H2O.
Then convert xx mols H2O to grams. g = mols x molar mass. I obtained approximately 9 g. This is the theoretical (100%) yield.
% yield if 5.4 g is obtained is
% yield = (5.4/approx 9)*100 =
mols WO3 = grams/molar mass = approx 0.17
Using the coefficients in the balanced equation, convert mols WO3 to mols H2O. That's ?mols WO3 x (3 mols H2O/1 mol WO3) = ?mols WO3 x 3 = xx mols H2O.
Then convert xx mols H2O to grams. g = mols x molar mass. I obtained approximately 9 g. This is the theoretical (100%) yield.
% yield if 5.4 g is obtained is
% yield = (5.4/approx 9)*100 =
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