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#??

Answer 1

#"117 g S"_8#

The idea here is that in order to identify whether or not you are working with a limiting reagent, you need to use the mole ratio that exists between the two reactants.

Examine the balanced chemical equation for this reaction first.

#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

Regarding sulfur dioxide, you'll need

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

Regarding hydrogen sulfide, you'll need

#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 can you determine if the reagent you're working with is limiting?

Using the mole ratios that exist between the two reactants, choose one reactant (sulfur dioxide, for example).

#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 precisely, the response will devour

#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 remainder of

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

will be excessive and, obliquely, not participate in the response.

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#

The number of sig figs you have for the samples of the two reactants, rounded to three, is what 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.
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Answer 2

First, balance the equation: 8SO2 + 16H2S -> 3S8 + 16H2O

Calculate the molar masses of each reactant and product: Molar mass SO2 = 32.07 g/mol + 215.999 g/mol = 64.07 g/mol Molar mass H2S = 21.008 g/mol + 32.07 g/mol = 34.08 g/mol Molar mass S8 = 8*32.07 g/mol = 256.56 g/mol

Calculate the number of moles of each reactant: Number of moles SO2 = 83.0 g / 64.07 g/mol = 1.296 mol Number of moles H2S = 83.0 g / 34.08 g/mol = 2.434 mol

Using the stoichiometry of the reaction, determine the limiting reactant: From the balanced equation, 8 moles of SO2 react with 16 moles of H2S to produce 3 moles of S8. So, the ratio of moles of SO2 to H2S is 1:2.

Since there are fewer moles of SO2 (1.296 mol) compared to H2S (2.434 mol), SO2 is the limiting reactant.

Calculate the maximum mass of S8 produced using the moles of the limiting reactant: 3 moles of S8 * 256.56 g/mol = 769.68 g

Therefore, the maximum mass of S8 that can be produced is 769.68 grams.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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