how many g of Fe2O3 are produced if 3.00 g of Fe reacts?
Fe = 56 g/mol
3 O = 3*16 = 48 g/mol
so a mol of Fe2O3 is 160 grams
and a mol of Fe is 56 grams
3 g Fe = 3/56 mol Fe
that will give me half as many mols of
Fe2O3
so
3/112 mols Fe2O3
(3/112) * 160 = 4.29 grams of Fe2O3
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To determine the amount of Fe2O3 produced when 3.00 g of Fe reacts, we need to first balance the chemical equation for the reaction. The balanced equation for the reaction between Fe and O2 to produce Fe2O3 is:
4 Fe + 3 O2 -> 2 Fe2O3
According to the balanced equation, we need 4 moles of Fe to produce 2 moles of Fe2O3. To find the moles of Fe in 3.00 g of Fe, we need to use the molar mass of Fe.
The molar mass of Fe is 55.85 g/mol. Therefore, the moles of Fe can be calculated as:
Moles of Fe = Mass of Fe / Molar mass of Fe
= 3.00 g / 55.85 g/mol
= 0.05365 mol
From the balanced equation, we know that 2 moles of Fe2O3 are produced for every 4 moles of Fe. Therefore, the moles of Fe2O3 can be calculated using the stoichiometry:
Moles of Fe2O3 = (Moles of Fe / 4) * 2
= (0.05365 mol / 4) * 2
= 0.02683 mol
Finally, to calculate the mass of Fe2O3 produced, we use the molar mass of Fe2O3, which is 159.69 g/mol:
Mass of Fe2O3 = Moles of Fe2O3 * Molar mass of Fe2O3
= 0.02683 mol * 159.69 g/mol
= 4.287 g
Therefore, 3.00 g of Fe will produce approximately 4.287 g of Fe2O3.