What is the maximum mass of #S_8# that can be produced by combining 83.0 g of each reactant in the reaction #8SO_2 + 16H_2S -> 3S_8 + 16H_2O#??

1 Answer
Mar 6, 2016


#"117 g S"_8#


The idea here is that you need to use the mole ratio that exists between the two reactants to determine whether or not you're dealing with a limiting reagent.

Start by taking a look at the balanced chemical equation for this reaction

#color(red)(8)"SO"_2 + color(blue)(16)"H"_2"S" -> 3"S"_8 + 16"H"_2"O"#

Sulfur dioxide, #"SO"_2#, and hydrogen sulfide, #"H"_2"S"#, will react in a #color(red)(1):color(blue)(2)# mole ratio, which means that the reaction will always consume twice as many moles of hydrogen sulfide as you have moles of sulfur dioxide taking part in the reaction.

Your next step will be to determine how many moles of each reactant you get in those #"83.0-g"# samples. To do that, use their respective molar masses

For sulfur dioxide you will have

#83.0 color(red)(cancel(color(black)("g"))) * "1 mole SO"_2/(64.064color(red)(cancel(color(black)("g")))) = "1.2956 moles SO"_2#

For hydrogen sulfide you will have

#83.0color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"S")/(34.081color(red)(cancel(color(black)("g")))) = "2.4354 moles H"_2"S"#

So, how do you test whether or not you're dealing with a limiting reagent?

Pick one reactant and use the mole ratios that exists between the two reactants. Let's go with sulfur dioxide here.

#1.2956color(red)(cancel(color(black)("moles SO"_2))) * (color(blue)(2)" moles H"_2"S")/(color(red)(1)color(red)(cancel(color(black)("mole SO"_2)))) = "2.5912 moles H"_2"S"#

So, in order for all the moles of sulfur dioxide to react, you need to have #2.5912# moles of hydrogen sulfide. SInce you have a little less than that, hydrogen sulfide will act as a limiting reagent here, i.e. it determine how many sulfur dioxide actually reacts.

More specifically, the reaction will consume

#2.4354color(red)(cancel(color(black)("moles H"_2"S"))) * (color(red)(1)" mole SO"_2)/(color(blue)(2)color(red)(cancel(color(black)("moles H"_2"S")))) = "1.2177 moles SO"_2#

The remaining

#1.2956 - 1.2177 = "0.0779 moles SO"_2#

will be in excess and, implicitly, not take part in the reaction.

Now that you know how many moles of each reactant take part in the reaction, pick one and use the mole ratio it has with sulfur, #"S"_8#, to determine how many grams would be theoretically result from the reaction.

Let's pick sulfur dioxide, which has a #color(red)(8):3# mole ratio with sulfur. If #1.2177# moles of sulfur dioxide react, the reaction will produce

#1.2177color(red)(cancel(color(black)("moles SO"_2))) * "3 moles S"_2/(color(red)(8)color(red)(cancel(color(black)("moles SO"_2)))) = "0.45664 moles S"_8#

To determine how many Grams of sulfur would contain that many moles, use the element's molar mass. Do not forget that you're dealing with molecular sulfur, hence the #8# subscript in #"S"_8#

#0.45664color(red)(cancel(color(black)("moles S"_8))) * (8 xx "32.065 g")/(1color(red)(cancel(color(black)("mole S"_8)))) = "117.14 g S"_8#

Rounded to three sig figs, the number of sig figs you have for the samples of the two reactants, the answer will be

#m_(S_8) = color(green)(|bar(ul(color(white)(a/a)"117 g S"_9color(white)(a/a)))|)#

This represents the theoretical yield of the reaction, which is calculated assuming a #100%# reaction yield.