Limiting reagent

[chiddler]

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The limiting reagent has a molecular mass of 128 g/mol and product has a molecular mass of 160 g/mol. If 0.2g of limiting reagent generates 0.2g of product, then what is percent yield?

Answer is 80%.

I'd really appreciate help with this as i'm having difficulty following TBR's math.

What i'm inefficiently doing:

128 to 160 represents approximately a 30% increase in mass. Ideally, 0.2 reagent should become 0.2 * 1.3 =0.26 but this is not observed. So 0.2 / 0.26=...

math gets too difficult. Using a calculator, I got 0.76.

So what's a good way to do this?

thank you.

 

[aSagacious]

Moles of limiting reagent used:
(0.2 g) / (128 g/mol) = 1.5625 x 10^-3 moles

Moles of product formed:
(0.2 g) / (160 g/mol) = 1.2500 x 10^-3 moles

...assuming the stoichiometry of the limiting reagent to the product is 1:1

((1.2500E-3) / (1.5625E-3)) x 100 = 80%

 

[MedPR]

The limiting reagent has a molecular mass of 128 g/mol and product has a molecular mass of 160 g/mol. If 0.2g of limiting reagent generates 0.2g of product, then what is percent yield?
.

Got halfway through my explanation then the library decided to close. Answer soon.

 

[BestDoctorEver]

128/160 = 0.8 since it's a 1 to 1 ratio limiting reagent to product. Why all the other calculations by other posters.

 

[chiddler]

sorry maybe I should have been more clear. i'm having difficulty following their logic. I don't need the solution because I already have it.

mr asagacious, your numbers are just as difficult as mine o_o

specifically,

((1.2500E-3) / (1.5625E-3)) x 100 = 80%

 

[MedPR]

sorry maybe I should have been more clear. i'm having difficulty following their logic. I don't need the solution because I already have it.

mr asagacious, your numbers are just as difficult as mine o_o

specifically,

((1.2500E-3) / (1.5625E-3)) x 100 = 80%

Well ((1.2500E-3) / (1.5625E-3)) is the same as 12/15, which is 0.8. But, the long way logic is this.

You start with ~16*10^-4mol of reactant. It is the limiting reactant, so theoretically all of it is consumed to form product. If it is a 1:1 ratio, you will form 16*10^-4mol of product, but your actual yield is only ~12*10^-4mol. So percent yield is theoretical/actual == 12*10^-4/16*10^-4 = ~80%.

 

[chiddler]

Well ((1.2500E-3) / (1.5625E-3)) is the same as 12/15, which is 0.8. But, the long way logic is this.

You start with ~16*10^-4mol of reactant. It is the limiting reactant, so theoretically all of it is consumed to form product. If it is a 1:1 ratio, you will form 16*10^-4mol of product, but your actual yield is only ~12*10^-4mol. So percent yield is theoretical/actual == 12*10^-4/16*10^-4 = ~80%.

75% (12/16) is one of the answer choices.

that said, i'm fine with the solutions you two gave me because they work with approximations. TBR's just being lame with difficult math.

 

[MedPR]

75% (12/16) is one of the answer choices.

that said, i'm fine with the solutions you two gave me because they work with approximations. TBR's just being lame with difficult math.

I think that's their little way of telling you to do 130/160 (the shortcut) instead.

 

[pfaction]

Can I ask why we're doing the 130/160? Is it because we'd only get 130g of product vs 160? Or what?

 

[MedPR]

Can I ask why we're doing the 130/160? Is it because we'd only get 130g of product vs 160? Or what?

R=reactant
P=product

0.2/128=molR
0.2/160=molP

error*molR=molP

(0.2/128)(error)=(0.2/160)

(0.2)(error)=(128)(0.2/160)

(0.2/0.2)(error)=(128/160)