Answer to Question #105649 in Chemistry for a

Question

What is the maximum mass of
S
8
that can be produced by combining
75.0
g
of each reactant?

8
SO
2
+
16
H
2
S
⟶
3
S
8
+
16
H
2
O

Answer 1

V=m/M

where m-mass, grams; M-molar mass, gram/mol.

M(SO2 )=64.1g/mol                        M(H2S)=34.1g/mol

ν(SO2) =84/64.1 =1.31 moles

ν(H2S) = 84/34.1 =2.46 moles

 calculating the amount of S8 , that can be produced from 84.0 grams of each reactant:

ν(S8 )=(1.31/8)*3= 0.49 moles

v(S8)=(2.46/16)*3=0.46 moles

As we can see from the previous calculations, the amount of H2S is the determining factor.

There will be an excess amount of S8. That is why:

m(s8)=v(s8)*M(s8)

M(S8 )=256.5 g/mol

That is why the maximum mass of S8, that can be produced is equal to:

m(S8)=0.46 *256.5=118g