How many grams of O2 are needed to produce 45.8 grams of Fe2O3 in the following reaction? 4Fe(s)+3O2(g) - 2Fe2O3(s)

Answer 1

Back track,

45.8 grams of Fe2O3*(1 mole/159.69 g)= moles of Fe2O3

moles of Fe2O3 *(3 moles of O2/2 moles of Fe2O3)=moles of O2

moles of O2*(32.0g of O2/1mole)= mass of O2

Answer 2

9.933 grams