2Ca(s)+O2(g)--->2CaO(s)
Calculate the mass of calcium oxide that can be prepared from 8.99 g of Ca and 5.99 g of O2.
To calculate the mass of calcium oxide (CaO) that can be prepared from the given amounts of calcium (Ca) and oxygen (O2), we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed in a chemical reaction, limiting the amount of product that can be formed.
To find the limiting reactant, we need to determine the number of moles of each reactant.
1. Calculate the number of moles of calcium (Ca):
- The molar mass of Ca is 40.08 g/mol.
- Divide the given mass of Ca (8.99 g) by the molar mass of Ca to get the number of moles:
8.99 g Ca / 40.08 g/mol = 0.224 mol Ca
2. Calculate the number of moles of oxygen (O2):
- The molar mass of O2 is 32.00 g/mol.
- Divide the given mass of O2 (5.99 g) by the molar mass of O2 to get the number of moles:
5.99 g O2 / 32.00 g/mol = 0.187 mol O2
3. Determine the mole ratio of Ca to CaO from the balanced chemical equation:
According to the balanced equation, 2 moles of Ca react with 1 mole of O2 to produce 2 moles of CaO.
4. Compare the mole ratios of Ca and O2:
The mole ratio of Ca to O2 is 2:1. However, the actual ratio of moles we calculated is:
- Ca: 0.224 mol
- O2: 0.187 mol
5. Identify the limiting reactant:
The limiting reactant is the one with the smallest mole ratio. In this case, oxygen (O2) has the smaller mole ratio, so it is the limiting reactant.
6. Use the mole ratio between CaO and O2 to calculate the theoretical yield of CaO:
According to the balanced equation, 2 moles of CaO are produced from 1 mole of O2. Since O2 is the limiting reactant, we will use its number of moles to calculate the theoretical yield.
- Number of moles of CaO = 1/2 * number of moles of O2
Number of moles of CaO = 1/2 * 0.187 mol O2 = 0.0935 mol CaO
7. Calculate the mass of CaO using the molar mass of CaO:
- The molar mass of CaO is 56.08 g/mol.
- Multiply the number of moles of CaO by its molar mass to get the mass:
Mass of CaO = 0.0935 mol CaO * 56.08 g/mol = 5.25 g
Therefore, the mass of calcium oxide that can be prepared from 8.99 g of Ca and 5.99 g of O2 is approximately 5.25 grams.
how many moles of Ca and O2 do you have?
Each mole of O2 uses 2 moles of Ca, so check to see which will run out first. Then change the moles of CaO back to grams.