Mole Calculation help! please! (1 Viewer)

[funkyknight]

hi,

please can you solve the following moles questions. I will post the solution that's in my text book but i don't understand it.

The rusting of iron can be represented by the following equation:
4Fe + 3O2 + 2H2O --> 2Fe2O3.H2O

How much iron rusts to produce 300.0g Fe2O3.H2O?

EDIT--> SOLUTION IN TEXTBOOK:
M = 2 x 55.85 + 3 x 16 + 2 x 1.008 + 16 = 177.716

n=m/M = 300.0/177.716 = 1.6881 moles Fe2O3.H2O

Moles Fe: n = 2 x moles Fe2O3.H2O = 2 x 1.6882 = 3.3762 moles Fe
Mass Fe: m = nM = 3.3762 x 55.85 = 188.6g

188.6g Iron rust..

Is this correct? Please help!

 

[Alkanes]

n(Fe2O3.H2O) = 300/177.716
= 1.688 mol

Ratio of Fe2O3H20 to Fe is 1:2

Therefore m(Fe) = 3.37 mol x 55.85g/mol
= 188.6g

 

[Alkanes]

You didn't answer the question. Lol. It asks how much iron rusts to produce the compound. Where did u get Al2O3 from?? :S

 

[funkyknight]

Haha I'm sorry! I posted the solution to the wrong question! Edit coming Soon