Question

Rust, chemically, is iron(III) oxide,Fe2O3.On a large ship, 16kg of rust accumulates in a day. Calculate how many kilograms of iron and how many kilograms of oxygen would be used up.

Answer 1

Fe=4.19
O=.9024

Answer 2

It's wrong.

Answer 3

hint: the molar mass of Fe2O3 is 159.70 So, the number of mol of Fe2O3 is:
n=16000/159.70= 100.18

Answer 4

now, 1 mol of Fe2O3 contains 2 moles of Fe and 3 moles of O, so we can write:
mFe=(100.18*2*55.85)/1000=11.2
mO=(100.18*3*16)/1000=4.8

Answer 5

where mO and mFe are the masses of oxygen and iron respectively

Answer 6

that's right! As we can see if we compute this sum: 11.2+4.8, we get the initial mass, namely 16 Kg