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

Answer:

$ions_{Fe^{2+}}=9.90x10^{22}ionsFe^{2+}$

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:

$ions_{Fe^{2+}}=25.0gFeSO_4*\frac{1molFeSO_4}{152gFeSO_4}*\frac{1molFe^{2+}}{1molFeSO_4}*\frac{6.022x10^{23}ionsFe^{2+}}{1molFe^{2+}} \\ions_{Fe^{2+}}=9.90x10^{22}ionsFe^{2+}$

Best regards.