1) Balance the chemical equation.
[tex]Fe+O_2\rightarrow Fe_2O_3[/tex]
Balance Fe.
[tex]2Fe+O_2\rightarrow Fe_2O_3[/tex]
Balance O.
[tex]4Fe+3O_2\rightarrow2Fe_2O_3[/tex]
2) Moles of Fe2O3 produced from 1.15 mol Fe.
The molar ratio between Fe2O3 and Fe is 2 mol Fe2O3: 4 mol Fe.
[tex]mol\text{ }Fe_2O_3=1.15\text{ }mol\text{ }Fe*\frac{2\text{ }mol\text{ }Fe_2O_3}{4\text{ }mol\text{ }Fe}=0.575\text{ }mol\text{ }Fe_2O_3[/tex]
1.15 has three significant figures. So, this result, 0.575 has three significant figures.
3) Convert moles of Fe2O3 to grams of Fe2O3
The molar mass of Fe2O3 is 159.6882 g/mol
[tex]g\text{ }Fe_2O_3=0.575\text{ }mol\text{ }Fe_2O_3*\frac{159.6882\text{ }g\text{ }Fe_2O_3}{1\text{ }mol\text{ }Fe_2O_3}=91.82\text{ }g\text{ }Fe_2O_3[/tex]
0.575 has three significant figures. So, this result, 91.82 must be rounded to three significant figures.
91.8 g Fe2O3 are produced from 1.15 mol Fe.
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