Question #051d8

1 Answer
Dec 8, 2015

Answer:

#94.7%#

Explanation:

Start from the balanced chemical equation

#color(blue)(2)"S"_text((s]) + color(red)(3)"O"_text(2(g]) -> 2"SO"_text(3(g])#

Notice the mole ratios that exist between the three chemical species that take part in the reaction.

You have #color(blue)(2)# moles of sulfur reacting with #color(red)(3)# moles of oxygen gas to produce #2# moles of sulfur trioxide. Your goal now is to use these mole ratios to determine the theoretical yield of the reaction.

Use sulfur and oxygen's molar masses to determine how many moles of each you have taking part in the reaction

#5.0 color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0 color(red)(cancel(color(black)("g")))) = "0.1563 moles O"_2#

#6.0 color(red)(cancel(color(black)("g"))) * "1 mole S"/(32.065 color(red)(cancel(color(black)("g")))) = "0.1871 moles S"#

At this point, you want to check if both reactants are completely consumed by the reaction, or if one of them will be a limiting reagent.

Pick one of the reactants, let's say sulfur, and use the #color(blue)(2) : color(red)(3)# mole ratio that exists between them to see how many moles of oxygen would be needed to ensure that all the moles of sulfur react

#0.1871 color(red)(cancel(color(black)("moles S"))) * (color(red)(3)" moles O"_2)/(color(blue)(2)color(red)(cancel(color(black)("moles S")))) = "0.2807moles O"_2#

So, in order for all the moles of sulfur to react, you would need to have #0.2807# moles of oxygen gas. Since you have fewer moles present, oxygen will be a limiting reagent.

Simply put, the oxygen gas will be consumed before all the sulfur can react.

#0.1563 color(red)(cancel(color(black)("moles O"_2))) * (color(blue)(2)" moles S")/(color(red)(3) color(red)(cancel(color(black)("moles O"_2)))) = "0.1042 moles S"#

Out of the available #0.1871# moles of sulfur, only #0.1042# moles will react. The rest will be in excess.

Now focus on finding the theoretical yield of the reaction. Since sulfur and sulfur trioxide have a #color(blue)(2):2# mole ratio, the number of moles of sulfur trioxide produced by the reaction will be equal to the number of moles of sulfur that actually react.

This means that you have

#0.1042 color(red)(cancel(color(black)("moles S"))) * "2 moles SO"_3/(color(blue)(2) color(red)(cancel(color(black)("moles S")))) = "0.1042 moles SO"_3#

Use sulfur trioxide's molar mass to determine how many Grams would contain this many moles

#0.1042 color(red)(cancel(color(black)("moles SO"_3))) * "80.06 g"/(1color(red)(cancel(color(black)("mole SO"_3)))) = "8.342 g"#

This will represent the theoretical yield of the reaction, i.e. how much sulfur trioxide will be produced if the reaction ahs a #100%# yield.

However, the reaction only produced #7.9# grams of sulfur, which means that its yield is actually

#color(blue)("% yield" = "actual yield"/"theoretical yield" xx 100)#

#"% yield" = (7.9 color(red)(cancel(color(black)("g"))))/(8.342 color(red)(cancel(color(black)("g")))) xx 100 = color(green)(94.7%)#

I'll leave the answer rounded to three sig figs, despite the fact that you only have two sig figs for the masses of the three species.