Find mass of Fe in `Fe_(3)O_(4)` containing `4.8xx10^(23)` oxygen atoms.

Options

A. 16.8 g
B. 33.6 g
C. 67.2 g
D. 48.2 g

Answer

B

Explanation

n of O = `(4.8xx10^(23))/(6xx10^(23))=0.8`
Let n of `Fe_(3)O_(4)impliesx`
`x xx 4 = 0.8`
`x = (0.8)/(4)=0.2 " mole of "Fe_(3)O_(4)`
mass of Fe = `0.2 xx 3 xx 56 = 33.6 g`