Question

Consider the following reaction:

2H2S (g) + 3O2 (g) --> 2SO2 (g) + 2H2O (g)

If O2 was the excess reagent, 1.3 moles of H2S were consumed, and 18.7 g of wat were collected after the reaction has gone to completion, what is the percent yield of the equation?

Answer 1

What is the theoretical yield?
1.3 moles H2S x (2 moles H2O/2 moles H2S) = 1.3 moles x (1/1) = 1.3 moles H2O produced. g H2O = moles x molar mass = 1.3 moles x 18 g/mol = about 23 estimated theoretical yield. Therefore, percent yield = (18.7/theoretical yield)*100 = ?

Answer 2

2H2S(g)+3O2(g)→2SO2(g)+2H2O(g)


A 1.2mol
sample of H2S(g)
is combined with excess O2(g)
, and the reaction goes to completion.

Question
Which of the following predicts the theoretical yield of SO2(g)
from the reaction?