Question #051d8
1 Answer
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
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
#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
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
Now focus on finding the theoretical yield of the reaction. Since sulfur and sulfur trioxide have a
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
However, the reaction only produced
#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.