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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?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» Correct Answer - B
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"`


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