how many grams of oxygen are in 213.5g Fe2O3?
213.5g Fe2O3 x (1 mole Fe2O3/molar mass Fe2O3) x (3 atoms oxygen/1 mole Fe2O3) x (16 g oxygen/1 mole O atoms) =
To find out how many grams of oxygen are in 213.5g of Fe2O3, you need to calculate the molar mass of Fe2O3 and determine the proportion of oxygen within the compound.
The molar mass of Fe2O3 can be determined by adding the atomic masses of its constituent elements: iron (Fe) and oxygen (O).
Fe: atomic mass = 55.845 g/mol
O: atomic mass = 16.00 g/mol
Since there are two atoms of iron (Fe) and three atoms of oxygen (O) in Fe2O3, you can calculate the molar mass as follows:
Molar mass of Fe2O3 = (2 * 55.845 g/mol) + (3 * 16.00 g/mol)
= 111.69 g/mol + 48.00 g/mol
= 159.69 g/mol
Now that you know the molar mass of Fe2O3, you can determine the proportion of oxygen in the compound. Oxygen contributes 48.00 g/mol to the total molar mass. So, the proportion of oxygen in Fe2O3 is:
Proportion of oxygen in Fe2O3 = (48.00 g/mol / 159.69 g/mol)
Lastly, to calculate the grams of oxygen, multiply the proportion by the given mass of Fe2O3:
Grams of oxygen = Proportion of oxygen in Fe2O3 * 213.5 g
Plug in the values:
Grams of oxygen = (48.00 g/mol / 159.69 g/mol) * 213.5 g
Now calculate the result to find out how many grams of oxygen are in 213.5g of Fe2O3.