Step 1 : Convert mass of Iron to moles using n = m/MM.
n(Fe) = 37.5 ÷ 55.845 = 0.671501... mol
Step 2 : Using stoichiometric ratios, find how many moles of Fe2O3 are produced.
Fe : Fe2O3 is 4:2, or simply 2:1. For every 2 moles of Fe there is Fe2O3.
Therefore, moles of Fe2O3 is half the amount of Fe!
n(Fe2O3) = 0.5 x 0.671501... = 0.33575... mol
Step 3: Convert the moles of Fe2CO3 to mass using m = n x MM.
m(Fe2O3) = 0.33575... x [2(55.845)+3(15.999)] = 53.61502... grams
Final step: round your answer to the lowest signficant figures. Since you gave me the mass of iron to 3s.f, m(Fe2O3) = <u><em>53.6 grams</em></u> (3s.f) !
Let me know if you're confused in any way!
