Question #3320d

1 Answer
P dilip_k
Oct 11, 2016

120g #S_8#

Explanation:

The equation of the given reaction is

#8SO_2+16H_2S->3S_8+16H_2O#

Considering the atomic masses of

#S->32" g/mol"#

#O->16" g/mol"#

#H->1" g/mol"#

We calculate the molar masses of

#SO_2->32+2*16=64" g/mol"#

#H_2S->2*1+32=34" g/mol"#

#S_8->8*32=256" g/mol"#

From the equation of the reaction we get the ratio of no. of moles of

#n_(SO_2):n_(H_2S):n_(S_8)=8:16:3#

So the ratio of their masses

#=m_(SO_2):m_(H_2S):m_(S_8)#

#=(8xx64):(16xx34):(3xx256)#

#=16:17:24=80:85:120#

The mass of each reactant is 85g .

The ratio of masses as shown above reveals that 85g of #H_2S# is exhausted when 80g of #SO_2# reacts with it and as a result maximum 120g #S_8# is produced.

Related questions
See all questions in Limiting Reagent
Impact of this question
1450 views around the world
You can reuse this answer
Creative Commons License