# percentage yield

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#1
A chemist mixed 12 g of phosphorus with 35.5 g of chlorine gas to synthesise phosphorous (iii) chloride. the yield was 42.4 g of PCl3. calculate the percentage yield for this reaction. the equation is:

2P + 3Cl2 --> 2PCl3

it'd be really helpful if someone could show me the calculations. thanks in advance 0
6 years ago
#2
(Original post by jana668)
A chemist mixed 12 g of phosphorus with 35.5 g of chlorine gas to synthesise phosphorous (iii) chloride. the yield was 42.4 g of PCl3. calculate the percentage yield for this reaction. the equation is:

2P + 3Cl2 --> 2PCl3

it'd be really helpful if someone could show me the calculations. thanks in advance Yes, this has a little more complication.

First off you have to determine the limiting reagent by calculating the moles of each reactant. Then you use the coefficients of the balanced equation (the balancing numbers) to work out which of the two reactants gets used up first.

So:

Moles of phosphorus = mass/Ar
Moles of chlorine = mass/Mr

Then you use the limiting reagent (the one that gets used up first) and the reaction coefficients to find out how many moles of product should be formed, this is the theoretical yield.

You then change moles to mass by: mass = mol x Mr

And this is the theoretical mass that should have been produced.

Finally you determine the percentage yield using:

percentage yield = 100 * actual yield/theoretical yield
0
#3
(Original post by charco)
Yes, this has a little more complication.

First off you have to determine the limiting reagent by calculating the moles of each reactant. Then you use the coefficients of the balanced equation (the balancing numbers) to work out which of the two reactants gets used up first.

So:

Moles of phosphorus = mass/Ar
Moles of chlorine = mass/Mr

Then you use the limiting reagent (the one that gets used up first) and the reaction coefficients to find out how many moles of product should be formed, this is the theoretical yield.

You then change moles to mass by: mass = mol x Mr

And this is the theoretical mass that should have been produced.

Finally you determine the percentage yield using:

percentage yield = 100 * actual yield/theoretical yield

when finding moles of chlorine, should i use 35.5 g for one mole or 71 g (coz its Cl2)? this is the actual doubt i have. can u tell me in what calculations we use the mass of 1 atom as the mass of one mole for diatomic molecules?
0
6 years ago
#4
(Original post by jana668)
when finding moles of chlorine, should i use 35.5 g for one mole or 71 g (coz its Cl2)? this is the actual doubt i have. can u tell me in what calculations we use the mass of 1 atom as the mass of one mole for diatomic molecules?
You always use the formula as shown in the balanced equation, so its Cl2
0
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