Yes, it is a limiting regent (LR) problem with and extra percent yield thrown in.
H2 + I2 ==> 2HI
mols H2 = grams/molar mass = ?
mols I2 = grams/molar mass = ?
Now, using the coefficients in the balanced equation, convert mols H2 to mols HI produced.
Do the same and convert mols I2 to mols HI. It is likely that the two values for mols HI will not agree. The correct answer in LR problems is ALWAYS the smaller value and the reagent responsible for that is the LR.
Now use the smaller value and convert mols HI to grams. grams HI = mols HI x molar mass HI. This is the theoretical yield (TY). The actual yield (AY) in the problem is stated as 43.1 grams HI.
%yield = (AY/TY)*100 = ?
What is the percent yield of the reaction below if 2.23g of hydrogen gas react with 55.3 g of iodine to produce 43.1g of HI. use the equation below:
H2+I2=2HI
the answers:
a) 21.8
b) 77.3
c) 33.7
d) 87.1
I started by trying to find the limiting reagent. I multiplied the 2.23 g of hydrogen by 2 moles of HI and got 4.46. Then I multiplied 55.3 g of I by 2 moles of HI and got 110.6. neither of those shows me which is the limiting reagent, and if that's the case, then if i did it wrong and was supposed to divide by the 2 moles, then it makes both of them the limiting reagent. Even if I could figure out the limiting reagent, then I don't know what to do next and am stuck.
3 answers
mols H2=1.106
mols I2=0.218
molar mass of H2=2.016 and divide 2.23g by 2.016
moles of H2 to moles of HI=2.212
moles of I2 to HI=0.436
so limiting reagent is I
grams of HI=511.6
(2*255.8)
percent yield=8.4%
(43.1/511.6)
What did I do wrong?
mols I2=0.218
molar mass of H2=2.016 and divide 2.23g by 2.016
moles of H2 to moles of HI=2.212
moles of I2 to HI=0.436
so limiting reagent is I
grams of HI=511.6
(2*255.8)
percent yield=8.4%
(43.1/511.6)
What did I do wrong?
Many of your numbers are right but they are in the wrong place. Here is what you should have done.
mols H2 = 2.23/2.016 = 1.106
mols I2 = 0.218. So these two are ok.
mols HI produced from H2 = 2*1.16 = 2.21
mols HI produced from I2 = 2*0.218 = 0.436 and you are right that I2 is the limiting reagent.
Here is your error.
grams HI = mols HI x molar mass HI.
g HI = 0.436 x 127.9 = 55.8 g HI. This is the theoretical yield (You used 2*HI)
%yield = (43.1/55.8)*100 = 77.2% and b is the closest answer.
mols H2 = 2.23/2.016 = 1.106
mols I2 = 0.218. So these two are ok.
mols HI produced from H2 = 2*1.16 = 2.21
mols HI produced from I2 = 2*0.218 = 0.436 and you are right that I2 is the limiting reagent.
Here is your error.
grams HI = mols HI x molar mass HI.
g HI = 0.436 x 127.9 = 55.8 g HI. This is the theoretical yield (You used 2*HI)
%yield = (43.1/55.8)*100 = 77.2% and b is the closest answer.
2 answers