The final mass of the iron bar and the rust when the weight of the bar was 664g and the bar had been standing in moist air for a month, will be 688.456g.
One-eighth of the iron bar turns to rust, which means the mass of the iron bar that has undergone rusting= 664/8= 83g
Thus, the mass of pure iron left after rusting has taken place will be= (664-83)g= 581g.
The equation of rusting is as follows
4Fe + 3O2 + 6H2O → 4Fe(OH)3
Molecular weight of iron is 55.84g, thus iron will undergo the reaction= 83/55.84= 1.486 mole
Mass of Fe2O3= 159.68g
4 moles of iron produce 4 moles of rust or iron oxide
Thus, 1.486 mole of iron will produce 1.486 mole iron oxide
Mass of iron oxide= 159.68/ 1.486
Mass of rust = 107.456g
Therefore, mass of iron bar= (581+ 107.456)g
= 688.456g
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