What is theoretical yield of Cr2O3 if the amount of Cr is 0.0382mol and the actual yield of Cr2O3 is 0.0186mol?
depends on the reaction equation.
Your post is confusing.
% yield = 100*(actual yield/theoretical yield)
theoretical yield = 100*actual yield/% yield
However, I don't know what that 0.0382 number is
Refer to this link above. That person had an equation that explained the question better. Here is the link.
https://www.jiskha.com/questions/1859998/what-is-theoretical-yield-of-cr2o3-if-the-amount-of-cr-is-0-0382mol-and-the-actual-yield
To find the theoretical yield of Cr2O3, we need to determine the stoichiometry of the reaction between Cr and O2 to produce Cr2O3. From the balanced chemical equation:
4 Cr + 3 O2 → 2 Cr2O3
We can see that 4 moles of Cr reacts with 3 moles of O2 to produce 2 moles of Cr2O3.
Given that the amount of Cr is 0.0382 mol, we can use the stoichiometry to calculate the theoretical yield of Cr2O3.
(0.0382 mol Cr) × (2 mol Cr2O3 / 4 mol Cr) = 0.0191 mol Cr2O3
So, the theoretical yield of Cr2O3 is 0.0191 mol.
The actual yield of Cr2O3 is given as 0.0186 mol. To calculate the percentage yield, we use the formula:
Percentage yield = (actual yield / theoretical yield) × 100
Percentage yield = (0.0186 mol / 0.0191 mol) × 100 = 97.9%
Therefore, the theoretical yield of Cr2O3 is 0.0191 mol, and the percentage yield is 97.9%.