2Ca(s) + O2(g) → 2CaO(s)
Calculate the mass of calcium oxide that can be prepared from 7.52 g of Ca and 5.01 g of O2.
oops!! oobleck misread the problem as CO2 instead of O2.
mols Ca = 7.52/40 = 0.188
mols O2 = 5.01/32 = 0.156
The problem tells you that 0.188 g Ca will use 0.094 mols O2 and produce 0.188 moles of CaO
0.188 mols CaO x 56 g CaO/mol CaO = 10.5 g CaO so the answer comes out the same although the moles O2 are not the same as moles CO2. Some people are just born lucky. BTW, the limiting reagent is Ca and the excess reagent is O2.
Calculate the mass of the quantity of the chemical in bold.
a 2Ca + O2 2CaO
(from 40 g of Ca) ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________
b 4Li + O2 2Li2O (from 30 g of Li)
______________________________________________________________________________
______________________________________________________________________________
______________________________________________________________________________
c Mg + 2HCl MgCl2 + H2 (from 85 g of MgCl2)
Well, well, well, looks like we have a little stoichiometry problem here. Don't worry, I'm here to calculate the mass of calcium oxide that can be formed from those ingredients.
First, let's find out which reactant is limiting the amount of calcium oxide we can make. We'll do this by calculating the number of moles of each reactant.
For calcium (Ca):
Molar mass of Ca = 40.08 g/mol
Number of moles of Ca = 7.52 g / 40.08 g/mol ≈ 0.1873 mol
For oxygen (O2):
Molar mass of O2 = 32.00 g/mol
Number of moles of O2 = 5.01 g / 32.00 g/mol ≈ 0.1566 mol
Now, let's look at the balanced equation: 2Ca(s) + O2(g) → 2CaO(s)
According to the equation, it takes 1 mole of O2 to react with 2 moles of Ca to produce 2 moles of CaO. So, the limiting reactant is oxygen (O2) because it has fewer moles. We'll use the number of moles of O2 to calculate the mass of CaO that can be formed.
Number of moles of CaO = 0.1566 mol × (2 mol CaO/1 mol O2) × (56.08 g/mol CaO)
Calculating that gives us approximately 17.6 grams of calcium oxide that can be produced.
So, from 7.52 g of calcium and 5.01 g of oxygen, it looks like we can make around 17.6 grams of calcium oxide. Keep in mind that this is the theoretical yield, meaning it's the maximum amount you can produce under ideal conditions.
To calculate the mass of calcium oxide (CaO) that can be prepared, we need to determine which reactant is limiting and then use stoichiometry to relate the limiting reactant to the desired product.
Step 1: Determine the limiting reactant
To do this, we need to compare the amount of each reactant to the stoichiometric ratio in the balanced equation.
The balanced equation is: 2Ca(s) + O2(g) → 2CaO(s)
The molar mass of Ca is 40.08 g/mol, and the molar mass of O2 is 32.00 g/mol.
First, calculate the number of moles of each reactant:
moles of Ca = mass of Ca / molar mass of Ca
moles of Ca = 7.52 g / 40.08 g/mol = 0.187 mol
moles of O2 = mass of O2 / molar mass of O2
moles of O2 = 5.01 g / 32.00 g/mol = 0.157 mol
Next, calculate the ratio of the reactants based on the stoichiometry of the balanced equation:
Ca:O2 = 2:1
So, for every 2 moles of Ca, we need 1 mole of O2. Therefore, if we have 0.187 mol of Ca, we need 0.0935 mol of O2. Since we only have 0.157 mol of O2 (which is greater than 0.0935 mol), O2 is in excess, and Ca is the limiting reactant.
Step 2: Calculate the mass of the product (CaO)
Using the stoichiometry of the balanced equation, we know that for every 2 moles of Ca, 2 moles of CaO are formed.
moles of CaO = moles of Ca
Therefore, the moles of CaO formed is 0.187 mol.
Finally, calculate the mass of CaO formed:
mass of CaO = moles of CaO × molar mass of CaO
mass of CaO = 0.187 mol × (40.08 g/mol)
mass of CaO = 7.5 g
Therefore, the mass of calcium oxide (CaO) that can be prepared from 7.52 g of Ca is approximately 7.5 g.
7.52g of Ca = 0.188 moles
5.01g of CO2 = 0.114 moles
The equation says that 0.188 moles of Ca will use 0.094 moles of CO2, and produce 0.188 moles of CaO = 10.54g