General Chemistry 1

Question

If 6.00 L of water vapor at 50.2 degrees celcius and 0.121 atm reacts with excess iron, how many grams of iron(III) oxide will be produced?

2Fe (s) + 3H2O (g) → Fe2O3 (s) + 3H2 (g)

Answer:

Mass= ______g

Answer 1

Hello, Madison.

We are provided a balanced equation and a limited volume of water vapor that reacts with excess iron. Since we know that water is the limiting reagent, we need the moles of water contained in the 6.00 liters of water vapor. Since the cnditions of this gas are not STP, we need the ideal gas law to calculate the number of moles of water.

PV=nRT

The volume and pressure can be left as if we use the value of the gas constant, R, that is 0.082058 L*atm*K^-1mole^-1. We do need to convert the temperature from C to K, by adding 273.15. Then solve the equation for n, the number nof moles of water contained in the 6.0 litersof water vapor at these conditions. I find 0.5681 moles H₂O.

The balanced equation tells us we only get 1 mole of Fe₂O₃ for every 3 moles of water. Meager. Since water is the limiting reagent, we'll assume all of it reacts to form Fe₂O₃.

(0.5681 moles H₂O)*(1 mole Fe₂O₃/3 moles H₂O) = 0.189 moles Fe₂O₃

Multiply by the molar mass of Fe₂O₃ (159.7 grams/mole) to find total mass of Fe₂O₃.

I find 30.2 grams Fe₂O₃ (3 sig figs). Wow. My car accumulates this amount of Fe₂O₃ in a day with far less effort.

Bob