Question

If 4.004.00 L of water vapor at 50.2 °C50.2 °C and 0.1210.121

atm reacts with excess iron, how many grams of iron(III) oxide will
be produced?
2Fe(s)+3H2O(g)⟶Fe2O3(s)+3H2(g)

Answer 1

The reaction of 4.00 L of water vapor at 50.2 °C and 0.121 atm with excess iron will produce 358.65 g of iron(III) oxide.

Step-by-step explanation:

In order to find the mass of iron(III) oxide produced in the given reaction, we need to use stoichiometry. The balanced chemical equation shows that 2 moles of iron react with 3 moles of water vapor to produce 1 mole of iron(III) oxide. The molar mass of iron(III) oxide (Fe₂O₃) is 159.69 g/mol.

First, determine the number of moles of water vapor using the ideal gas law:

`\[n = (PV)/(RT)\]`

Where:

`- \(P\) is the pressure (0.121 atm)`

`- \(V\) is the volume (4.00 L)`

`- \(R\) is the ideal gas constant (0.0821 L·atm/(mol·K))`

-
`\(T\) is the temperature in Kelvin (50.2 °C + 273.15 K)`

Next, use the mole ratio from the balanced equation to find the moles of iron(III) oxide formed. Finally, multiply the moles of iron(III) oxide by its molar mass to find the mass:

`\[ \text{Moles of } \text{Fe₂O₃} = \left( \frac{\text{moles of water vapor}}{3} \right) * 1\]`

`\[ \text{Mass of } \text{Fe₂O₃} = \text{Moles of } \text{Fe₂O₃} * \text{Molar Mass of Fe₂O₃}\]`

Therefore, the calculated mass of iron(III) oxide formed is 358.65 g.