When 28g calcium reacts with 25 grams oxygen, how many grams calcium oxide form?
I have 2Ca+ 2O2= 4CaO its is not working. Stoichemistry. I have the Ca as the reactant that is limiting? Please help.
Your equation is not balanced. I'll bet if you balanced it the stoichiometry would work. If it still doesn't work post you work and I'll find the error.
To determine which reactant is limiting, you need to compare the moles of each reactant to their respective stoichiometric coefficients.
1. Calculate the number of moles for each reactant:
- Moles of calcium (Ca):
Moles = mass / molar mass = 28 g / 40.08 g/mol = 0.6999 mol (rounded to four decimal places)
- Moles of oxygen (O2):
Moles = mass / molar mass = 25 g / 32 g/mol = 0.7813 mol (rounded to four decimal places)
2. Compare the mole ratios from the balanced equation to determine the limiting reactant.
From the balanced equation: 2Ca + 2O2 -> 4CaO
The mole ratio of Ca to CaO is 2:4 or 1:2.
The mole ratio of O2 to CaO is 2:4 or 1:2.
Calculating the maximum moles of CaO that can be formed:
- If Ca is the limiting reactant: 0.6999 mol Ca * (2 mol CaO / 2 mol Ca) = 0.6999 mol CaO
- If O2 is the limiting reactant: 0.7813 mol O2 * (4 mol CaO / 2 mol O2) = 1.5626 mol CaO
Since the calculated moles of CaO from the limiting reactant are lower, Ca is the limiting reactant.
3. Calculate the number of grams of calcium oxide (CaO) formed:
Moles of CaO = 0.6999 mol (from step 2)
Mass = Moles x Molar mass
Mass = 0.6999 mol x 56.08 g/mol = 39.16 g (rounded to two decimal places)
Therefore, when 28g of calcium reacts with 25 grams of oxygen, approximately 39.16 grams of calcium oxide will form.
To solve this stoichiometry problem, you need to determine which reactant is the limiting reagent.
First, find the moles of each reactant. The molar mass of calcium (Ca) is 40.08 g/mol, and the molar mass of oxygen (O) is 16.00 g/mol.
1. Moles of calcium (Ca):
Moles = mass / molar mass
Moles of Ca = 28 g / 40.08 g/mol
2. Moles of oxygen (O):
Moles = mass / molar mass
Moles of O = 25 g / 16.00 g/mol
Next, use the balanced equation to determine the stoichiometric ratio between calcium and oxygen. The balanced equation you provided is correct: 2 Ca + 2 O2 -> 4 CaO
The ratio is 2:2, which means that for every 2 moles of calcium, 2 moles of oxygen are needed to produce 4 moles of calcium oxide (CaO).
Now, compare the moles of calcium and oxygen to determine the limiting reagent. The limiting reagent is the reactant that is the least in amount, as it will be completely consumed in the reaction.
3. Moles of calcium oxide (CaO) that can be formed from calcium:
Moles of CaO = (moles of Ca) / (ratio of Ca to CaO)
Moles of CaO from Ca = (28 g / 40.08 g/mol) / (2 mol Ca / 4 mol CaO)
4. Moles of calcium oxide (CaO) that can be formed from oxygen:
Moles of CaO = (moles of O) / (ratio of O to CaO)
Moles of CaO from O = (25 g / 16.00 g/mol) / (2 mol O / 4 mol CaO)
Compare the moles of CaO from calcium and oxygen, and the smaller value will indicate the limiting reagent. The limiting reagent determines the maximum amount of product that can be formed.
Finally, convert the moles of calcium oxide obtained from the limiting reagent into grams by multiplying by the molar mass of calcium oxide (CaO), which is 56.08 g/mol.
5. Grams of calcium oxide (CaO):
Grams = moles × molar mass
Grams of CaO = (moles of CaO) × (molar mass of CaO)
This process should help you calculate the correct amount of calcium oxide that will form in the reaction.