Question

If you started with 35.0 grams of H2S and 40.0 grams of O2, how many grams of S8 would be produced, assuming 95 % yield? 8H2S(g)+4O2(g)→S8(l)+8H2O(g)

Answer 1

Molar mass

H2S = 34.0 g/mol
O2 = 31.99 g/mol
S8 = 256.52 g/mol

Identifying excess reagent and the limiting of the reaction :

8 H2S(g) + 4 O2(g) = S8(I) +  8 H2O(g)

8 x 34 g H2S -------->  256. 52 g S8
35.0 g ----------------> ??

35.0 x 256.52 / 8 x 34 =

8978.2 / 272 => 33.00 g of S8  

H2S is  the  limiting reactant
---------------------------------------------

4 x 31.99 g O2 --------------- 256.52 g S8
40.0 g O2 --------------------- ??

40.0 x 256.52 / 4 x 31.99 =

10260.8 / 127.96 = 80.16 g of S8  

O2 is the excess reagent is the excess reagent

------------------------------------------------------------

H2S is the limiting reactant, one that is fully consumed, it is he who determines the mass of S8 produced 

33.0 g ----------- 100%
?? g ------------- 95 %

95 x 33.00 / 100 => 31.35 g

hope this helps!