What is the mass of 5.33 mol Fe2(SO4)3?
Answer in units of g
Look on the periodic table and add them up.
Take the mass of Iron and multiply it by 2, then get the mass of Sulfur and multiply it by 3, then get the mass of Oxygen and multiply it by 12 and then add them up. Once you have that mass, then you have its "molar mass" so just multiply the molar mass by the amount of mols!
To find the mass of 5.33 mol of Fe2(SO4)3, we need to use the concept of molar mass. The molar mass is the mass of one mole of a substance, expressed in grams.
First, we need to calculate the molar mass of Fe2(SO4)3. To do this, we sum up the atomic masses of all the elements in the compound. Here are the atomic masses of the elements involved:
- Atomic mass of Fe (iron) = 55.845 g/mol
- Atomic mass of S (sulfur) = 32.06 g/mol
- Atomic mass of O (oxygen) = 16.00 g/mol
Now let's calculate the molar mass of Fe2(SO4)3:
Molar mass of Fe2(SO4)3 = (2 x Atomic mass of Fe) + (3 x Atomic mass of S) + (12 x Atomic mass of O)
= (2 x 55.845) + (3 x 32.06) + (12 x 16.00) g/mol
= 111.690 + 96.180 + 192.00 g/mol
= 399.870 g/mol
Therefore, the molar mass of Fe2(SO4)3 is approximately 399.870 g/mol.
To find the mass of 5.33 mol of Fe2(SO4)3, we can multiply the molar mass by the number of moles:
Mass = Number of moles x Molar mass
= 5.33 mol x 399.870 g/mol
Calculating this:
Mass = 2132.757 g
So, the mass of 5.33 mol Fe2(SO4)3 is approximately 2132.757 g.