In the reaction #4Fe + 3O_2 -> 2Fe_2O_3#, how many grams of #Fe_2O_3# are produced from #73.2g# #Fe#?

Answer 1

Avery Adams

104.6 g of #"Fe"_2"O"_3#.

A solution to this can be found via dimensional analysis.

Masses of Atoms:

#"Fe = 55.85 g/mol"# #"O = 16.00 g/mol"#

Formula Weight:

#"Fe"_2"O"_3 ="159.69 g/mol"#

#"Moles of Fe" =73.2cancel("g Fe") xx ("1 mol Fe")/(55.85cancel("g Fe")) = "1.31 mol Fe"#

#"Moles of Fe"_2"O"_3 =1.31 cancel("mol Fe") xx ("2 mol Fe"_2"O"_3)/(4cancel("mol Fe")) = "0.655 mol Fe"_2"O"_3#

#"Mass of Fe"_2"O"_3 = 0.655cancel("mol Fe"_2"O"_3) xx ("159.69 g Fe"_2"O"_3)/(1 cancel("mol Fe"_2"O"_3)) = "104.6 g Fe"_2"O"_3#

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Answer 2

Cooper Anderson

To solve this problem, first, calculate the molar mass of Fe and Fe2O3. Then, use stoichiometry to find the amount of Fe2O3 produced from 73.2g of Fe.

Calculate the molar mass of Fe: Fe: 55.845 g/mol
Calculate the molar mass of Fe2O3: Fe2O3: (2 * 55.845 g/mol) + (3 * 16.00 g/mol) = 159.69 g/mol
Use the stoichiometry of the reaction: 4Fe + 3O2 -> 2Fe2O3
Calculate the moles of Fe using its molar mass: Moles of Fe = (73.2 g Fe) / (55.845 g/mol) = 1.311 moles Fe
Use the stoichiometry to find the moles of Fe2O3 produced: Moles of Fe2O3 = (1.311 moles Fe) * (2 moles Fe2O3 / 4 moles Fe) = 0.6555 moles Fe2O3
Convert moles of Fe2O3 to grams using its molar mass: Mass of Fe2O3 = (0.6555 moles Fe2O3) * (159.69 g/mol) = 104.53 g Fe2O3

So, 73.2 grams of Fe will produce 104.53 grams of Fe2O3.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

In the reaction #4Fe + 3O_2 -&gt; 2Fe_2O_3#, how many grams of #Fe_2O_3# are produced from #73.2g# #Fe#?

In the reaction #4Fe + 3O_2 -> 2Fe_2O_3#, how many grams of #Fe_2O_3# are produced from #73.2g# #Fe#?