calculate the number of oxygen needed to oxidize 12.5grams of glucose (C6H12O6) into carbon dioxide and water. (Combustion reaction)
first we write the combustion reaction:
C6H12O6 + O2 -> H2O + CO2
then we balance it:
C6H12O6 + 6O2 -> 6H2O + 6CO2
since mass of glucose is given, we get its molecular weight,, from the periodic table mass of each element in glucose is
H = 1 , C = 12 , O = 16
6*12 + 12*1 + 6*16 = 180 g/mol
thus,
12.5 g C6H12O6 * (1 mol C6H12O6 / 180 g C6H12O6) * (6 mol O2 / 1 mol C6H12O6) = 0.4167 mol O2
or in grams, (O2 = 32 g/mol)
32*0.4167 = 13.33 g O2
hope this helps~ :)
Ah, the magic of combustion! Now, let me put on my clown hat and bring some humor to this calculation.
To calculate the number of oxygen atoms needed, let's break down the atomic compositions of glucose (C6H12O6) and carbon dioxide (CO2):
1 molecule of glucose (C6H12O6) yields 6 molecules of carbon dioxide (6CO2) and 6 molecules of water (6H2O).
Now, let's look at the atomic ratios. In one molecule of glucose (C6H12O6), we have:
- 6 carbon atoms (C)
- 12 hydrogen atoms (H)
- 6 oxygen atoms (O)
And for every molecule of carbon dioxide (CO2), we have:
- 1 carbon atom (C)
- 2 oxygen atoms (O)
Since 6 molecules of carbon dioxide are produced per molecule of glucose, we multiply the number of carbon atoms (6C) and oxygen atoms (6O) by 6.
Therefore, we need:
- 6 carbon atoms x 6 molecules = 36 carbon atoms (C)
- 6 oxygen atoms x 6 molecules = 36 oxygen atoms (O)
But wait, we also have to consider the oxygen atoms that come from glucose itself.
One glucose molecule already contains 6 oxygen atoms (6O). Since we're dealing with 12.5 grams of glucose, we'll need to calculate the moles using the molar mass of glucose. The molar mass of glucose (C6H12O6) is approximately 180 g/mol.
So, 12.5 g of glucose is:
12.5 g / 180 g/mol ≈ 0.0694 mol
Since there are 6 oxygen atoms in one glucose molecule, we multiply 0.0694 mol by 6 to get the total number of oxygen atoms needed:
0.0694 mol x 6 = approximately 0.4165 mol
So, in conclusion, you'd need approximately 0.4165 moles of oxygen to oxidize 12.5 grams of glucose into carbon dioxide and water. Keep in mind that this calculation assumes the reaction goes to completion, and we rounded the moles for simplicity.
But hey, remember that behind this scientific calculation, there's always a clown trying to make you smile!
To calculate the number of oxygen molecules needed to oxidize glucose (C6H12O6) into carbon dioxide (CO2) and water (H2O) in a combustion reaction, we can follow these steps:
Step 1: Write the balanced chemical equation for the combustion of glucose:
C6H12O6 + 6O2 → 6CO2 + 6H2O
Step 2: Determine the molar mass of glucose (C6H12O6):
C = 6(12.01 g/mol) = 72.06 g/mol
H = 12(1.01 g/mol) = 12.12 g/mol
O = 6(16.00 g/mol) = 96.00 g/mol
Total = 72.06 + 12.12 + 96.00 = 180.18 g/mol
Step 3: Calculate the number of moles of glucose:
Number of moles = mass / molar mass
Number of moles = 12.5 g / 180.18 g/mol ≈ 0.069 moles
Step 4: Since the balanced equation shows that 1 mole of glucose reacts with 6 moles of O2, we can calculate the number of moles of O2 needed:
Number of moles of O2 = 0.069 moles of glucose × 6 moles of O2 / 1 mole of glucose = 0.414 moles of O2
Step 5: Convert the number of moles of O2 to molecules:
Number of molecules = number of moles × Avogadro's number
Number of molecules = 0.414 moles × 6.022 × 10^23 molecules/mol ≈ 2.49 × 10^23 molecules of O2
Therefore, to oxidize 12.5 grams of glucose into carbon dioxide and water, approximately 2.49 × 10^23 molecules of oxygen are needed.
To calculate the number of oxygen molecules required to oxidize glucose, we need to first understand the balanced equation for the combustion of glucose.
The balanced equation for the combustion of glucose (C6H12O6) can be represented as follows:
C6H12O6 + 6O2 -> 6CO2 + 6H2O
From the balanced equation, we can see that one molecule of glucose reacts with six molecules of oxygen to produce six molecules of carbon dioxide and six molecules of water.
To determine the amount of oxygen needed to completely oxidize 12.5 grams of glucose, we will follow these steps:
Step 1: Calculate the molar mass of glucose (C6H12O6).
The molar mass of carbon (C) is approximately 12 g/mol, hydrogen (H) is approximately 1 g/mol, and oxygen (O) is approximately 16 g/mol. Therefore, the molar mass of glucose can be calculated as follows:
(6 * 12 g/mol) + (12 * 1 g/mol) + (6 * 16 g/mol) = 72 g/mol + 12 g/mol + 96 g/mol = 180 g/mol
Step 2: Convert the mass of glucose (12.5 grams) to moles by dividing it by the molar mass.
12.5 grams / 180 g/mol = 0.06944 moles
Step 3: Use the stoichiometry of the balanced equation to calculate the number of moles of oxygen required.
From the balanced equation, we know that the molar ratio of glucose to oxygen is 1:6 (6 moles of O2 are needed for each mole of glucose).
Therefore, the number of moles of oxygen required can be calculated as follows:
0.06944 moles * 6 = 0.4167 moles
Step 4: Convert the number of moles of oxygen to molecules.
Since there are approximately 6.022 x 10^23 molecules in one mole, we can calculate the number of oxygen molecules required as follows:
0.4167 moles * 6.022 x 10^23 molecules/mol = 2.506 x 10^23 molecules
Therefore, approximately 2.506 x 10^23 molecules of oxygen are needed to completely oxidize 12.5 grams of glucose into carbon dioxide and water.