How many grams of Fe2O3 are formed when 51.3 grams of iron, Fe, react completely with oxygen, O2? 4 Fe + 3 O2 --> 2 Fe2O3

The Fe is our limiting reactant here. First, convert the mass of Fe given to moles of Fe by dividing mass by the molar mass:

(51.3 g Fe)/(55.845 g/mol) = 0.9186 mol Fe.

According to the balanced equation, 2 moles of Fe2O3 are produced for every 4 moles of Fe that are consumed. In other words, half as many moles of Fe2O3 will be produced as there are Fe that react:

(0.9186 mol Fe)(2 mol Fe2O3/4 mol Fe) = 0.4593 mol Fe2O3.

Finally, convert the moles of Fe2O3 back to grams by multiplying by the molar mass of Fe2O3:

(0.4593 mol Fe2O3)(159.6882 g/mol) = 73.3 g Fe2O3.

So, 73.3 grams of Fe2O3 are formed.

it is reliable if two scientists recorded it