What maximum mass of sulphur in kilograms could be produced?

Sulphur is used in the production of vulcanized rubber, detergents, dyes, pharmaceuticals, insecticides, and many other substances. The final step in the Claus method for making sulphur is #SO_2(g)+2H_2S(g) rarr 3S(s)+2H_2O(l)#
The first question asked to find the minimum volume of sulphur dioxide at STP that would be necessary to produce 82.5 kg of sulphur. I was able to calculate that #1.92*10^4 L SO_2# was necessary.
The second question is the one I am stuck at. What maximum mass of sulphur in kilograms could be produced by the reaction of #(1.3*10^4 )m^3# of #SO_2(g)# with #(2.5*10^4)m^3# of #H_2S(g)#, both at STP?
Maybe I am just overlooking something obvious, but I feel a bit lost with problem.
Thank you very much in advance.

1 Answer
Meave60
Jul 14, 2018

Question 2

The maximum mass of sulfur that can be produced from the given volumes of #"SO"_2"# and #"H"_2"S"# is #5.5xx10^4# #"kg"#.

Explanation:

Balanced equation

#"SO"_2("g")+ "2H"_2"S(g)"##rarr##"3S(s) + 2H"_2"O("l")"#

Question 1: Volume of #"82.5 kg SO"_2"#

#82500color(red)cancel(color(black)("g S"))xx(1"mol S")/(32.06color(red)cancel(color(black)("g S")))="2573 mol S"#

#2573color(red)cancel(color(black)("mol S"))xx(1"mol SO"_2)/(3color(red)cancel(color(black)("mol S")))="857.7 mol SO"_2"#

At STP of #0^@"C"# and #"1 atm"#, the molar volume of a gas is #"22.414 L/mol"#. To calculate the volume of #"857.7 mol SO"_2"#, multiply mol #"SO"_2"# by the molar volume of a gas.

#857.7color(red)cancel(color(black)("mol SO"_2))xx("22.414 L")/(1color(red)cancel(color(black)("mol")))=1.92xx10^4# #"L SO"_2"#

Question 2: Maximum mass of sulfur in kilograms

Part 1) #1.3xx10^4# #"m"^3"# #"SO"_2"#

Convert #"m"^3"# to #"L"#.

#"1 m"^3"##=##"1000 L"#

#1.3xx10^4color(red)cancel(color(black)("m"^3))"SO"_2xx("1000 L")/(1color(red)cancel(color(black)("m"^3)))=1.3xx10^7# #"L SO"_2"#

We can determine moles #"SO"_2"# using the molar volume. Multiply the volume in liters by the inverse of the molar volume.

#1.3xx10^7color(red)cancel(color(black)("L")) "SO"_2xx(1"mol")/(22.414color(red)cancel(color(black)("L")))=5.8xx10^5# #"mol SO"_2"#

Determine moles #"S"# by multiplying mol #"SO"_2"# by the mole ratio between #"S"# and #"SO"_2"# in the balanced equation, with #"S"# in the numerator.

#5.8xx10^5color(red)cancel(color(black)("mol SO"_2))xx(3"mol S")/(1color(red)cancel(color(black)("mol SO"_2")))=1.7xx10^6# #"mol S"#

Determine the mass of #"S"# by multiplying mol #"S"# by its molar mass #("32.06 g/mol")#.

#1.7xx10^6color(red)cancel(color(black)("mol S"))xx(32.06"g S")/(1color(red)cancel(color(black)("mol S")))=5.5xx10^7# #"g S"#

Convert mass in grams to kilograms.

#5.5xx10^7color(red)cancel(color(black)("g")) "S"xx("1 kg")/(1000color(red)cancel(color(black)("g")))=5.5xx10^4# #"kg S"#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Part 2) #2.5xx10^4# #"m"^3" H"_2"S"#

Convert #"m"^3"# to #"L"#.

#2.5xx10^4color(red)cancel(color(black)("m"^3)) "H"_2"S"xx("1000 L")/(1color(red)cancel(color(black)("m"^3")))=2.5xx10^7# #"L H"_2"S"#

To get the moles of #"H"_2"S"#, multiply the volume in liters by the inverse of the molar volume.

#2.5xx10^7color(red)cancel(color(black)("L")) "H"_2"S"xx(1"mol")/(22.414color(red)cancel(color(black)("L")))=1.1xx10^6# #"mol H"_2"S"#

Determine moles #"S"# by multiplying mol #"H"_2"S"# by the mole ratio between #"S"# and #"H"_2"S"# in the balanced equation, with #"S"# in the numerator.

#1.1xx10^6color(red)cancel(color(black)("mol H"_2"S"))xx(3"mol S")/(2color(red)cancel(color(black)("mol H"_2"S")))=1.7xx10^6# #"mol S"#

Determine the mass of #"S"# by multiplying mol #"S"# by its molar mass #("32.06 g/mol")#.

#1.7xx10^6color(red)cancel(color(black)("mol S"))xx(32.06"g S")/(1color(red)cancel(color(black)("mol S")))=5.5xx10^7# #"g S"#

Convert mass in grams to kilograms.

#5.5xx10^7color(red)cancel(color(black)("g")) "S"xx(1"kg")/(1000color(red)cancel(color(black)("g")))=5.5xx10^4# #"kg S"#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The maximum mass of sulfur that can be produced from the given volumes of #"SO"_2"# and #"H"_2"S"# is #5.5xx10^4# #"kg"#.

Related questions
Impact of this question
4733 views around the world
You can reuse this answer
Creative Commons License