Determine the number of moles of sulfate ions present in 1.20 mol iron(III) sulfate (Fe2(SO4)3).

Would you believe 1.20 mol Iron(III))sulfate x (3 sulfate ions/1 mol iron(III) sulfate) = 3.60

3.6

To determine the number of moles of sulfate ions present in 1.20 mol of iron(III) sulfate (Fe2(SO4)3), we need to consider the chemical formula of iron(III) sulfate.

The formula Fe2(SO4)3 indicates that there are two iron (Fe) atoms and three sulfate (SO4) groups in each molecule of iron(III) sulfate. Each sulfate group consists of one sulfur (S) atom and four oxygen (O) atoms.

From the formula, we can see that there are three sulfate ions (SO4) per one molecule of iron(III) sulfate (Fe2(SO4)3). Therefore, to calculate the number of moles of sulfate ions, we need to multiply the number of moles of iron(III) sulfate by the ratio of sulfate ions to iron(III) sulfate.

The ratio of sulfate ions to iron(III) sulfate is 3:1. This means that for every 1 mole of iron(III) sulfate, there are 3 moles of sulfate ions. Therefore, we can multiply the number of moles of iron(III) sulfate by 3 to find the number of moles of sulfate ions.

In this case, we have 1.20 moles of iron(III) sulfate. Multiplying this by 3, we get:

1.20 mol x 3 = 3.60 moles of sulfate ions

Therefore, there are 3.60 moles of sulfate ions present in 1.20 moles of iron(III) sulfate.