Quicklime (CaO) can be prepared by roasting
limestone (CaCO3) according to the reaction
CaCO3(s)
�
−!CaO(s) + CO2(g) .
When 2.6 ×10^3 g of CaCO3 are heated, the actual yield of CaO is 1.06 ×10^3 g. What is the percent yield?
Answer in units of %.
To find the percent yield, we need to compare the actual yield to the theoretical yield.
The balanced equation for the reaction shows that 1 mole of CaCO3 produces 1 mole of CaO. Therefore, the molar mass of CaCO3 (100.09 g/mol) is equal to the molar mass of CaO (56.08 g/mol).
First, let's calculate the theoretical yield of CaO based on the given mass of CaCO3:
Moles of CaCO3 = mass of CaCO3 / molar mass of CaCO3
Moles of CaCO3 = 2.6 × 10^3 g / 100.09 g/mol
Moles of CaCO3 = 25.98 mol
Since the balanced equation shows a 1:1 molar ratio between CaCO3 and CaO, the moles of CaO produced will also be 25.98 mol.
The theoretical yield of CaO can be calculated as follows:
Theoretical yield = moles of CaO × molar mass of CaO
Theoretical yield = 25.98 mol × 56.08 g/mol
Theoretical yield = 1.459 × 10^3 g
Now, we can calculate the percent yield using the actual yield and theoretical yield:
Percent yield = (actual yield / theoretical yield) × 100
Percent yield = (1.06 × 10^3 g / 1.459 × 10^3 g) × 100
Percent yield = 72.78%
Therefore, the percent yield is approximately 72.78%.
To find the percent yield, we need to compare the actual yield to the theoretical yield. The theoretical yield is the maximum amount of product that can be obtained based on the stoichiometry of the reaction.
First, we need to calculate the molar mass of CaCO3 and CaO:
- The molar mass of CaCO3 (calcium carbonate) is: 40.08 g/mol (atomic mass of Ca) + 12.01 g/mol (atomic mass of C) + (3 * 16.00 g/mol) (atomic mass of three O) = 100.09 g/mol.
- The molar mass of CaO (calcium oxide or quicklime) is: 40.08 g/mol (atomic mass of Ca) + 16.00 g/mol (atomic mass of O) = 56.08 g/mol.
Now, we can calculate the theoretical yield of CaO using stoichiometry:
- From the balanced equation, we see that the mole ratio of CaCO3 to CaO is 1:1. This means that 1 mole of CaCO3 will produce 1 mole of CaO.
- Convert the mass of CaCO3 to moles using its molar mass:
Moles of CaCO3 = mass / molar mass
Moles of CaCO3 = 2.6 × 10^3 g / 100.09 g/mol = 25.97 mol
- Since the mole ratio is 1:1, the moles of CaO produced will be the same as the moles of CaCO3 used.
- Convert moles of CaO to mass using its molar mass:
Mass of CaO = moles of CaO * molar mass
Mass of CaO = 25.97 mol * 56.08 g/mol = 1.46 × 10^3 g
Now, we can calculate the percent yield:
- Percent Yield = (Actual Yield / Theoretical Yield) * 100
- Percent Yield = (1.06 × 10^3 g / 1.46 × 10^3 g) * 100
- Percent Yield = 72.6%
Therefore, the percent yield of CaO is 72.6%.
You must find the theoretical yield. Follow the steps in this link to find that. Tha link also hows how to calculate %yield.
http://www.jiskha.com/science/chemistry/stoichiometry.html