A mixture contains 0.250 mol of Mn2O3 and 20.0 g of MnO2.
1.How many atoms of oxygen are present in the mixture?
2.How many grams of manganese are in the mixture?
1. Well, to find the number of oxygen atoms, we first need to calculate the number of moles of oxygen in each compound. Mn2O3 has a 2:3 mole ratio of oxygen to manganese, so there are 0.250 * 3 = 0.750 moles of oxygen atoms. MnO2 has a 2:2 mole ratio of oxygen to manganese, so there are 20.0 / 86.94 * 2 = 0.458 moles of oxygen atoms. Adding these together, we get a total of 0.750 + 0.458 = 1.208 moles of oxygen atoms. Now, if we multiply this by Avogadro's number (6.022 x 10^23), we find that there are approximately 7.3 x 10^23 oxygen atoms in the mixture. That's a whole lot of atoms!
2. Now, for the grams of manganese in the mixture, we need to calculate the grams of each compound separately. Mn2O3 has a molar mass of 158.88 g/mol, so there are 0.250 * 158.88 = 39.72 grams of manganese in Mn2O3. MnO2 has a molar mass of 86.94 g/mol, so there are 20.0 grams of manganese in MnO2. Adding these together, we get a total of 39.72 + 20.0 = 59.72 grams of manganese in the mixture.
To answer these questions, we will first calculate the number of moles of MnO2, then use this information to calculate the number of atoms of oxygen and grams of manganese in the mixture.
1. Number of atoms of oxygen:
First, we need to calculate the number of moles of oxygen in Mn2O3 and MnO2.
Mn2O3: The formula Mn2O3 tells us that there are 3 oxygen atoms for every 1 molecule of Mn2O3.
Therefore, the number of moles of oxygen in Mn2O3 is 0.250 mol multiplied by 3, which equals 0.750 mol.
MnO2: The atomic weight of oxygen is 16.00 g/mol.
The number of moles of oxygen in MnO2 can be calculated using the given mass of MnO2, which is 20.0 g.
We can convert this mass to moles using the molar mass of MnO2.
The molar mass of MnO2 is the sum of the atomic masses of manganese (Mn) and two oxygen (O) atoms:
1 mol Mn = 54.94 g/mol
2 mol O = 2 * 16.00 g/mol = 32.00 g/mol
Molar mass of MnO2 = 54.94 g/mol + 32.00 g/mol = 86.94 g/mol
Now, we can calculate the number of moles of oxygen in MnO2 using the formula:
Number of moles = Given mass / Molar mass
Number of moles of oxygen in MnO2 = 20.0 g / 86.94 g/mol ≈ 0.230 mol
Finally, we add the moles of oxygen from Mn2O3 and MnO2 to get the total number of moles of oxygen:
Total moles of oxygen = moles of oxygen from Mn2O3 + moles of oxygen from MnO2 = 0.750 mol + 0.230 mol = 0.980 mol
To calculate the number of atoms of oxygen, we need to multiply the total moles of oxygen by Avogadro's number (6.022 x 10^23 atoms/mol):
Number of atoms of oxygen = Total moles of oxygen * Avogadro's number = 0.980 mol * (6.022 x 10^23 atoms/mol) ≈ 5.89 x 10^23 atoms
Therefore, there are approximately 5.89 x 10^23 atoms of oxygen in the mixture.
2. Grams of manganese:
To calculate the grams of manganese (Mn) in the mixture, we need to consider the moles of Mn2O3 and MnO2.
Mn2O3: The formula Mn2O3 tells us that there are 2 manganese atoms for every 1 molecule of Mn2O3.
Therefore, the number of moles of manganese in Mn2O3 is 0.250 mol multiplied by 2, which equals 0.500 mol.
MnO2: The atomic weight of manganese is 54.94 g/mol.
The number of moles of manganese in MnO2 can be calculated using the given mass of MnO2, which is 20.0 g.
We can convert this mass to moles using the molar mass of MnO2:
Number of moles of Mn in MnO2 = Given mass / Molar mass
Number of moles of Mn in MnO2 = 20.0 g / 86.94 g/mol ≈ 0.230 mol
Finally, we add the moles of manganese from Mn2O3 and MnO2 to get the total number of moles of manganese:
Total moles of manganese = moles of manganese from Mn2O3 + moles of manganese from MnO2 = 0.500 mol + 0.230 mol = 0.730 mol
To calculate the grams of manganese, we need to multiply the total moles of manganese by the molar mass of manganese:
Grams of manganese = Total moles of manganese * Molar mass of manganese = 0.730 mol * 54.94 g/mol ≈ 40.1 g
Therefore, there are approximately 40.1 grams of manganese in the mixture.
To solve these questions, we need to use the concept of moles and the Avogadro's number.
1. To find the number of atoms of oxygen present in the mixture, we need to calculate the number of moles of oxygen in each compound and then add them together.
For Mn2O3:
- Molecular formula: Mn2O3
- Oxygen atoms per formula unit: 3
- Total moles of oxygen in Mn2O3: 0.250 mol x 3 = 0.750 mol
For MnO2:
- Molecular formula: MnO2
- Oxygen atoms per formula unit: 2
- Total moles of oxygen in MnO2: (20.0 g / molar mass of MnO2) x (1 mol / 32.00 g) x 2 = 1.250 mol
Total moles of oxygen in the mixture = 0.750 mol + 1.250 mol = 2.000 mol
Now, we can calculate the number of atoms of oxygen:
Number of atoms of oxygen = Total moles of oxygen x Avogadro's number
= 2.000 mol x 6.022 x 10^23 atoms/mol
= 1.2044 x 10^24 atoms
Therefore, there are approximately 1.2044 x 10^24 oxygen atoms present in the mixture.
2. To find the grams of manganese in the mixture, we need to calculate the mass of each compound and then add them together.
For Mn2O3:
- Molecular formula: Mn2O3
- Molecular mass: (2 x atomic mass of Mn) + (3 x atomic mass of O)
- Molecular mass of Mn2O3: (2 x 54.94 g/mol) + (3 x 16.00 g/mol) = 228.84 g/mol
- Mass of Mn2O3 in the mixture: 0.250 mol x 228.84 g/mol = 57.21 g
For MnO2:
- Molecular formula: MnO2
- Molecular mass: atomic mass of Mn + (2 x atomic mass of O)
- Molecular mass of MnO2: 54.94 g/mol + (2 x 16.00 g/mol) = 86.94 g/mol
- Mass of MnO2 in the mixture: 20.0 g
Total mass of manganese in the mixture = Mass of Mn2O3 + Mass of MnO2
= 57.21 g + 20.0 g
= 77.21 g
Therefore, there are approximately 77.21 grams of manganese in the mixture.
1. use moles instead of grams for MnO2
(0.250*3 + 20.0/86.94 * 2) * 6.023*10^23
2. use grams instead of moles for Mn2O3 ...