Question

How many grams of iron oxide, Fe2O3 will be produced if 165 g of O2 gas is supplied? (follow the same steps as mol to mol, only now your flow should be like this: grams O2 moles O2  moles Fe2O3  grams Fe2O3 Fe + O2  Fe2O3

Answer 1

`m_(Fe_2O_3)=549gFe_2O_3`

Step-by-step explanation:

Hello there!

In this case, according to the given chemical reaction for this problem about stoichiometry:

`4Fe+3O_2\rightarrow 2Fe_2O_3`

Whereas there is a 3:2 mole ratio of oxygen (molar mass = 32.0 g/mol) to iron (III) oxide (molar mass = 159.69 g/mol) and therefore, the correct stoichiometric setup is:

`m_(Fe_2O_3)=165gO_2*(1molO_2)/(32.00gO_2)*(2molFe_2O_3)/(3molO_2) *(159.69gFe_2O_3)/(1molFe_2O_3) \\\\m_(Fe_2O_3)=549gFe_2O_3`

Regards!