If 64.0 g of H2O are actually obtained, find the actual yield %.
Show that the % the same whether you calculate using grams or moles.
Theoretical yield is 4.04 moles and in grams is 72.80 grams
2H2 + 1O2=2H2O
% yield = (actual/theoretical)*100 = ?
with mols:
mols H2O in 64g is 64/18 = 3.555
%yield = (3.55/4.04)*100 = 88.0%
with grams:
%yield = (64.0/72.8)*100 = 87.91; however, if I remember this problem right the number was 72.7 g H2O (g = mols x molar mass = 4.04 x 18 = 72.7) and
(64.0/72.7)*100 = 88.0%
You get 88.0% either way.
To find the actual yield percentage, we need to divide the actual yield by the theoretical yield, and then multiply by 100.
First, let's calculate the actual yield percentage using grams.
Actual yield = 64.0 g
Theoretical yield = 72.80 g
Actual yield percentage = (Actual yield / Theoretical yield) x 100
Plugging in the values, we get:
Actual yield percentage = (64.0 g / 72.80 g) x 100 ≈ 87.90%
Now, let's calculate the actual yield percentage using moles.
Actual yield = 64.0 g
Theoretical yield = 4.04 moles
To convert grams to moles, we need to use the molar mass of water (H2O), which is 18.02 g/mol.
Actual yield in moles = Actual yield / Molar mass of water
= 64.0 g / 18.02 g/mol
≈ 3.55 moles
Actual yield percentage = (Actual yield / Theoretical yield) x 100
Plugging in the values, we get:
Actual yield percentage = (3.55 moles / 4.04 moles) x 100 ≈ 87.87%
We can see that the actual yield percentage calculated using grams and moles is approximately the same (87.90% and 87.87% respectively). This shows that the percentage is independent of whether we calculate using grams or moles, as long as the correct conversions are made.