Can someone please help me.....
I have limited reactant formula problem. I am completely lost.
2Ca(s) + O2(g) ----> 2CaO(s)
Calculate the mass of Calcium Oxide that can be prepared from 6.87g of Ca and 4.58g of O2.
I calculated 2Ca to .171mol and O2 to .286mol, I am lost from here...
This is a limiting reagent (LR) problem and I work these the long way and probably not at all like your prof does. But here goes.
mols Ca = 6.87/atomic mass Ca = approx 0.171 which you have.
mols O2 = 4.58/32 = approx 0.143 ( you used the atomic mass of O and not the molar mass of O2).
The next step is to convert mols Ca to mols CaO AND convert mols O2 to mols CaO.
First Ca to CaO is 0.171 x (2 mol CaO/2 mol CaO) = 0.171 mols CaO formed IF we had all of the O2 we needed.
Next O2 to CaO is 0.143 x (2 mols CaO/1 mol O2) = 0.286 mols CaO IF we had all of the Ca we needed.
These two values are the same so one of them must be wrong. The correct value in LR problems is ALWAYS the smaller value so the correct value is we form 0.171 mols CaO and Ca is the limiting reagent.
Now use the smaller value and grams CaO = mols x molar mass.
Sweet. Thank you. I was real close. I was just calc ing the moles wrong at the end.
Thanks again
To solve this problem, you need to determine which reactant is the limiting reactant. The limiting reactant is the one that is completely consumed and determines the maximum amount of product that can be obtained.
Let's calculate the number of moles of calcium oxide that can be produced from each reactant:
From 6.87g of Ca:
Using the molar mass of Ca: 40.08 g/mol
Number of moles of Ca = mass of Ca / molar mass of Ca
= 6.87 g / 40.08 g/mol
≈ 0.171 mol
From 4.58g of O2:
Using the molar mass of O2: 32.00 g/mol
Number of moles of O2 = mass of O2 / molar mass of O2
= 4.58 g / 32.00 g/mol
≈ 0.143 mol
Now, we need to determine the limiting reactant. The balanced equation tells us that the stoichiometric ratio between Ca and CaO is 2:2, and the stoichiometric ratio between O2 and CaO is 1:2.
Since the stoichiometric ratio between Ca and CaO is 2:2, and we have 0.171 mol of Ca, we can say that 0.171 mol of Ca will produce 0.171 mol of CaO.
Since the stoichiometric ratio between O2 and CaO is 1:2, and we have 0.143 mol of O2, we can say that 0.143 mol of O2 will produce 0.286 mol of CaO.
Comparing the moles of CaO produced from each reactant, we see that 0.171 mol is less than 0.286 mol. Therefore, Ca is the limiting reactant.
Now, let's calculate the mass of calcium oxide produced from the limiting reactant:
Using the molar mass of CaO: 56.08 g/mol
Mass of CaO produced = moles of CaO x molar mass of CaO
= 0.171 mol x 56.08 g/mol
≈ 9.59 g
Therefore, the mass of calcium oxide that can be prepared from 6.87g of Ca and 4.58g of O2 is approximately 9.59g.
To solve this problem, you need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed, thereby limiting the amount of product that can be formed.
First, calculate the molar mass of Ca and O2:
- The molar mass of Ca is 40.08 g/mol.
- The molar mass of O2 is 32.00 g/mol.
Next, use the molar mass to convert the given masses of Ca and O2 to moles:
- For Ca: 6.87 g Ca * (1 mol Ca / 40.08 g Ca) ≈ 0.171 mol Ca
- For O2: 4.58 g O2 * (1 mol O2 / 32.00 g O2) ≈ 0.143 mol O2
Now, we can compare the moles of Ca and O2 to the balanced equation to determine the limiting reactant. The coefficients in the balanced equation tell us the mole ratio of reactants. From the balanced equation, we see that the stoichiometric ratio of Ca to O2 is 2:1. This means that for every 2 moles of Ca, we need 1 mole of O2.
To find out how much CaO can be formed, we can compare the moles of Ca and O2. Since the mole ratio of Ca to O2 is 2:1, for every 0.143 mol of O2, we will need 2 * 0.143 mol of Ca. This is equal to 0.286 mol of Ca.
Now, compare the actual amount of Ca you have (0.171 mol) with the required amount (0.286 mol). You can see that the amount of Ca you have is less than what is required (0.171 mol < 0.286 mol), which means that Ca is the limiting reactant.
Since the balanced equation tells us that 2 moles of Ca gives us 2 moles of CaO, we can assume that the moles of CaO formed will be equal to the moles of Ca (0.171 mol).
Finally, calculate the mass of CaO formed using the molar mass of CaO (56.08 g/mol):
- Mass of CaO = 0.171 mol CaO * (56.08 g CaO / 1 mol CaO) ≈ 9.60 g CaO.
Therefore, the mass of Calcium Oxide that can be prepared from 6.87g of Ca and 4.58g of O2 is approximately 9.60 grams.