n the manufacturing process of sulfuric acid, sulfur dioxide is reacted with oxygen to produce sulfur trioxide. Using the equation, 2SO2 (g) + O2 arrow 2SO3 (g), if 128 g of sulfur dioxide is given the opportunity to react with an excess of oxygen, but only produces 144 g of sulfur trioxide, what is the percent yield of this reaction?

Question

anonymous · Answer

@Zeta

anonymous · Answer

I know you can use the limiting reagent to solve it...is sulfur dioxide the limiting reagent? If so....eh, I have no idea what the hell I'm doing :(

anonymous · Answer

If you have 128 grams of sulphur dioxide, 

number of moles of sulphur dioxide = 128/(32 + 32) = 2 moles

Your reaction is $\rm 2 SO_2 + O_2 \to 2SO_3$, so the expected yield on the right hand side is 2 moles.

2 moles of $\rm SO_3$ = 2 * (molar mass of SO3) = 2 * (32 + 48) = 2 * 80 = 160 grams.

So you would expect 160 grams of sulphur trioxide, right?

anonymous · Answer

Sulphur dioxide is the limiting reagent here, but we don't need to think about limiting reagents here.

anonymous · Answer

makes sense...mostly
how did you get  again?
sorry, ust don't want to keep asking the same kinds of questions over and over

anonymous · Answer

get *32
sorry, lag