A mixture contains 0.149 mol of Mn2O3 and 14.9 g of MnO2.
and?
And what does that have to do with the study of Boise state university? If the university is your course of study, or school subject, I'm sure it's interesting and worth studying, but is that chemistry?
To find the total amount of manganese in the mixture, we need to calculate the number of moles of Mn2O3 and MnO2 separately.
Let's start with Mn2O3. Given that the mixture contains 0.149 mol of Mn2O3, we can find the molar mass of Mn2O3 using the periodic table. The molar mass of Mn is approximately 54.94 g/mol, and each mole of Mn2O3 contains two moles of Mn. Therefore, the molar mass of Mn2O3 is:
2 * 54.94 + 16.00 = 111.88 g/mol
To find the total mass of Mn2O3 in the mixture, we multiply the number of moles (0.149 mol) by the molar mass (111.88 g/mol):
0.149 mol * 111.88 g/mol = 16.69 g
Now let's move on to MnO2. Given that the mixture contains 14.9 g of MnO2, we can find the number of moles of MnO2 by dividing the mass by the molar mass. The molar mass of MnO2 is:
54.94 + 32.00 = 86.94 g/mol
To find the number of moles, we divide the mass (14.9 g) by the molar mass (86.94 g/mol):
14.9 g / 86.94 g/mol ≈ 0.171 mol
Now that we have the number of moles for each compound, we can find the total amount of manganese in the mixture by adding the moles of Mn from Mn2O3 and MnO2:
0.149 mol + 0.171 mol = 0.32 mol
Therefore, the total amount of manganese in the mixture is 0.32 moles.