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) = 53.6 grams (3s.f) !
Let me know if you're confused in any way!
Grams are the unit of the mass that is used to calculate the moles. From 37.5 gms of iron, 53.6 gms of ferric oxide are produced.
Mass is the measurement of the moles of the substance and the molar mass.
Moles of iron from the mass is calculated as:
Moles of iron = 37.5 gms ÷ 55.84 = 0.671 moles
The balanced chemical reaction:
4Fe + 3O2 → 2Fe2O3
From the above it is deduced that 4 moles of iron produce 2 moles of ferric oxide so, 0.671 moles of iron will produce,
(0.671 × 2) ÷ 4 = 0.3375 moles
Mass of ferric oxide, from moles, is calculated as:
Mass = 0.33 moles × 159.687
= 52.696 gms
Therefore, 53.6 gms of ferric oxide will be produced from 37.5 gms of iron.
Learn more about mass here:
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