well, let’s use stoichiometry to find out. 30.01g Fe2O3 x (1molFe203/159.69gFeNo3) * (4molFe/2molFeNO3) x (55.845gFe/1molFe) = 20.99 g Fe needed to produce 30.01g of Fe2O3 So, this is true!
Confused where I got the values from? - the 30.01g Fe2O3 comes from the problem. - the 159.69g Fe2NO3 is the molar mass found by added the mass of each element and how much of each element is present. -the 4mol of Fe and the 2mol Fe2NO3 are found in the balanced equation. -the 55.845 g of Fe is the molar mass of fe on the periodic table. hope this helps!
