Question

How many fe(ii) ions are there in 25.0 g of feso4?

Answer 1

Hey there!:

1 mole of FeSO4 = 151.8 g/mol

25.0 g change to mole = 25.0 / 151.8

moles of FeSO4 = 0.165 moles

One mole of any substance has 6.02*10²³ units

0.165 moles to get the number of Fe ( II ) ions :

number of Fe(II) ion = 0.165 * ( 6.02*10²³ )

number of Fe ( II ) ion = 9.93*10²² units

Answer 2

`ions_(Fe^(2+))=9.90x10^(22)ionsFe^(2+)`

Step-by-step explanation:

Hello,

In this case, the following mole-mass relationship, is applied to obtain the ions of iron (II) in 25.0 g of iron (II) sulfate whose molar mass is 152 g/mol:

`ions_(Fe^(2+))=25.0gFeSO_4*(1molFeSO_4)/(152gFeSO_4)*(1molFe^(2+))/(1molFeSO_4)*(6.022x10^(23)ionsFe^(2+))/(1molFe^(2+)) \\ions_(Fe^(2+))=9.90x10^(22)ionsFe^(2+)`

Best regards.