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

Answer:

$m_{Fe_2O_3}=549gFe_2O_3$

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:

$m_{Fe_2O_3}=165gO_2*\frac{1molO_2}{32.00gO_2}*\frac{2molFe_2O_3}{3molO_2} *\frac{159.69gFe_2O_3}{1molFe_2O_3} \\\\m_{Fe_2O_3}=549gFe_2O_3$

Regards!