65 g of iron rusts over time. Assumin no elemental iron (Fe) is left un oxidized how many grams of rust Fe2O3

are present??

4Fe+3O2---->2Fe2O3

Last one sorry for all the questions. You've help me so much!

It's just stoichiometry :)
molar mass of Fe= 55.85 g/mol
65g Fe/55.85g = 1.16 moles Fe
For every 4 moles Fe consumed, 2 moles of Fe2O3 are produced.
So 1.16 moles Fe/ 4 moles Fe x 2 moles Fe2O3 = 0.582 moles Fe2O3 are produced.
The molar mass of Fe2O3 is 159.7g/mol so just use this to convert from moles to grams...
0.582 moles Fe2O3 x 159.7g = 92.9 grams of rust Fe2O3 are present

This question says that the iron rusts (or oxidizes) completely. Therefore there is just enough oxygen to bond with the 65 grams of iron present.

Convert 65 grams of iron to moles of iron by multiplying by the inverted molar mass of iron (in units of mol/g). Next find the molar mass of iron oxide (2 times molar mass of iron + 3 times molar mass of oxygen). Then use stoichiometry to determine how many moles of iron oxide you would get from completely rusting the 65 grams of iron and then multiply this by the molar mass of iron oxide.

65 g Fe * ( 1 mol Fe / 55.842 g Fe ) * ( 2 mol Iron Oxide / 4 mol Fe ) * ( ( 2 * 55.842 g Fe + 3 * 16 g O ) / 1 mol Iron Oxide )

Answer: ( 65 * (55.842/1) * (4/2) * ( (2*55.842 + 3*16) / 1) ) grams of Iron Oxide.

You can calculate it fromthe atomic weights
Fe = 55.845
O = 15.9994
2 atoms of Fe = 2*55.845 = 111.69
3 atoms of O = 3*15.9994 = 47.9982
Then for 65 g of Fe we get 65 + 47.9982*65/111.69 g of Fe2O3
= 92.93 g of Fe2O3