How many moles of oxygen gas are consumed in the production of 5.00 g of iron(III) oxide from metallic iron? 4Fe(s)+3O2(g)→2Fe2O3(g) mole(s) O2

0.0469 mol of 02.

Explanation:

The balanced equation of this reaction is given as follows:

4Fe(s) + 3O2(g) → 2Fe2O3(s)

According to this equation, 3 moles of oxygen gas (O2) is consumed to produce 2 moles of iron(III) oxide (Fe2O3).

Using the formula, mole = mass/molar mass to convert mass of Fe2O3 to mole

Molar mass of Fe2O3 = 56(2) + 16(3)

= 112 + 48

= 160g/mol

mole = 5.00 g ÷ 160g/mol

mole = 0.03125mol of Fe2O3

Since 3 moles of oxygen gas (O2) is consumed to produce 2 moles of iron(III) oxide (Fe2O3).

0.03125mol of Fe2O3 will be produced when (0.03125 × 3/2) mol of oxygen gas is consumed.

0.03125 × 3/2

0.03125 × 1.5

= 0.0469 mol of 02.

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