You must first determine the theoretical yield. Here is an example. Just follow the steps. http://www.jiskha.com/science/chemistry/stoichiometry.html
The %yield = (actual yield/theoretical yield)*100 = ??
CaO(s) + H2O(l) �¨ Ca(OH)2(s)
In a particular experiment, a 2.00-g sample of CaO is reacted with excess water and 2.14 g of Ca(OH)2 is recovered. What is the percent yield in this experiment?
The %yield = (actual yield/theoretical yield)*100 = ??
Step 1: Calculate the molar mass of Ca(OH)2.
Ca: 1 mol × 40.08 g/mol = 40.08 g/mol
O: 2 mol × 16.00 g/mol = 32.00 g/mol
H: 2 mol × 1.01 g/mol = 2.02 g/mol
Total: 74.10 g/mol
Step 2: Calculate the theoretical yield of Ca(OH)2.
First, we need to calculate the number of moles of Ca(OH)2 produced using the given mass of CaO.
Molar mass of CaO (CaO): 1 mol × 40.08 g/mol + 1 mol × 16.00 g/mol = 56.08 g/mol
Number of moles of CaO: 2.00 g CaO × (1 mol CaO / 56.08 g CaO) = 0.0356 mol CaO
From the balanced equation, we know that 1 mol of CaO produces 1 mol of Ca(OH)2.
So, the number of moles of Ca(OH)2 produced is also 0.0356 mol.
Theoretical yield of Ca(OH)2: 0.0356 mol × 74.10 g/mol = 2.6416 g
Step 3: Calculate the percent yield.
Percent yield = (actual yield / theoretical yield) × 100
Percent yield = (2.14 g / 2.6416 g) × 100 ≈ 81.1%
Therefore, the percent yield in this experiment is approximately 81.1%.
First, you need to determine the theoretical yield of Ca(OH)2. This can be calculated from the balanced equation. The molar ratio between CaO and Ca(OH)2 is 1:1, which means that for every 1 mol of CaO reacted, 1 mol of Ca(OH)2 is produced.
1 mol of CaO has a molar mass of 56.08 g/mol, so the number of moles of CaO can be calculated by dividing the mass of CaO by its molar mass:
2.00 g CaO / 56.08 g/mol CaO = 0.0356 mol CaO
Since the molar ratio between CaO and Ca(OH)2 is 1:1, the number of moles of Ca(OH)2 produced is also 0.0356 mol.
The molar mass of Ca(OH)2 is 74.09 g/mol, so the theoretical yield of Ca(OH)2 can be calculated by multiplying the number of moles by its molar mass:
0.0356 mol Ca(OH)2 × 74.09 g/mol Ca(OH)2 = 2.64 g Ca(OH)2
Next, you can calculate the percent yield by dividing the actual yield (2.14 g Ca(OH)2) by the theoretical yield (2.64 g Ca(OH)2) and multiplying by 100:
Percent Yield = (Actual Yield / Theoretical Yield) × 100
Percent Yield = (2.14 g / 2.64 g) × 100 = 81.06%
Therefore, the percent yield in this experiment is approximately 81.06%.