Consider the following reaction: 2H2S(g)+3O2(g) yields 2SO2(g) +2H2O(g). If O2 was the excess reagent 8.3 mol of H2S were consumed and 137.1 g of...

To solve a problem like this, look at the coefficients of
the balanced chemical equation.  In this case, it tells you that for every two moles of
H2S you react, you will produce two moles of HOH.  That assumes a 100%
yield.

In the given problem, since you started with 8.3
moles of H2S, you would expect 8.3 moles of HOH at 100%
yield.

In your case, you produced 137.1 g of water. 
Convert this to moles which means divide 137.1 by the formula mass of water (18.016
g/mole) .

137.1 g/ 18.016 g/mole = 7.61 moles of
water.

To find % yield, divide the actual yield by the
theoretical yield and multiply by 100.

% yield = 7.61/8.3 *
100 = 91.69%