chemistry

Question

Determine the mass of iron (III) oxide produced when 2.50 moles of iron reacts with4.22 moles of oxygen. show all work

Positivelife · Answer

4 Fe (s)        +     3 O2(g)              -------.>            2 Fe2O3 (s)
  
Molar ratio :                 4                 :        3                             :                2   
No. of moles/mol         2.50                  4.22
  
 
  
Here, the limiting reagent is Fe.
  
It can be found like this,
  
Let's imagine the limiting reagent is O2,   
then the no. of moles of Fe should be reacted = (4.22 mol / 3) * 4
  
                                                                                      =  5.63 mol
  
Therefore, 5.63 mol of Fe should be reacted with O2 to finish O2.
  
But, the no. of moles of Fe is less than that amount. So that, it completely reacts with O2, remaining O2 in excess.
  
So, the no. of moles of Fe2O3 produced is totally depend on the no. of moles of Fe.
  
The no. of moles of Fe2O3 formed                     = (2.50 mol /4 )* 2
  
                                                                                    = 1.25 mol
  
Relative molecular mass of Fe2O3                      = (56 g mol^-1 *2) + (16 g mol^-1 * 3)
  
                                                                                    = 160 g mol^-1
  
Therefore, the mass of Fe2O3 Produced                       = 160 g mol^-1 * 1.25 mol
  
                                                                                    = 200 g

Manish411 · Answer

you can check calculations in attachment which is attached.

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