Question

What is the maximum mass of S8 that can be produced by combining 77.0 g of each reactant?

8SO2 + 16H2S ----> S8 + 16H2O

Answer 1

Answer;

=98.98g of S8

Solution and explanation;

The equation for the reaction is;

8 SO2 + 16 H2S → 3 S8 + 16 H2O

Therefore;

Moles of SO2

(77.0 g SO2) / (64 g SO2/mol) = 1.203 moles

Moles of H2S

(77.0 g H2S) / (34 g H2S/mol) = 2.26 moles

2.26 moles of H2S would react completely with 2.26 x (3/16) = 0.424 moles SO2, but there is more SO2 present than that, which means that SO2 is in excess and H2S is the limiting reactant.

1.203 moles SO2 yields 3/8x amount that is = 0.456moles of S8 , thus

2.06moles H2S yields 3/16x amount = 0.386moles of S8

Hence ;

H2S will produce only 0.386moles of S8 that is equivalent to 98.98g of S8

Answer 2

The balanced equation is
8SO2 + 16H2S ----> 3S8 + 16H2O
so
77g SO2 / 64g/mole = 1.203moles of SO2
77g H2S / 34g/mole = 2.26moles of H2S
now

1.203 moles SO2 yields 3/8x amount that is = 0.456moles of S8
so
2.06moles H2S yields 3/16x amount = 0.386moles of S8
hence
H2S = produce only 0.386moles of S8 = 98.98g S8
hope it helps