How many moles of ferric alum
`(NH_(4))_(2) SO_(4) Fe_(2) (SO_(4))_(3). 24 H_(2)O` can be made from the sample of `Fe` containing `0.0056 g` of it?

Options

A. `10^(-4) mol`
B. `0.5 xx 10^(-4) mol`
C. `0.33 xx 10^(-4) mol`
D. `2 xx 10^(-4) mol`

Answer

B

Explanation

Moles of `Fe = (0.0056)/(56) = 10^(-4) "mol"`
1 mol of alum = 2 mol of `Fe`
2 mol of `Fe = 1` mol of alum
`10^(-4) "mol of" Fe (1)/(2) xx 10^(-4) "mol"`
`= 0.5 xx 10^(-4) "mol"`