Question

What mass of Fe is required to react with 455 mL of Oxygen at 22 degree Celsius and a pressure of 725 torr according to the following equation. 4 Fe(s) + 3 O2(g) arrow 2 FeO3 (s).

Answer 1

In this question, we are going to use the Ideal Gas Law, which is a gas law that perfectly correlates temperature, pressure and volume of a gas, and we will be looking for the number of moles of O2 in this reaction, the formula is:

PV = nRT

Where:

P = pressure in atm, 0.95 atm

V = volume in Liters, 0.455 L

n = number of moles

R = gas constant, 0.082 L*atm/mol*K

T = temperature in Kelvin, 295 K

Now we add these values into the formula

0.95*0.455 = n*0.082*295

0.432 = 24.19n

n = 0.018 moles of O2

Now that we have the number of moles of O2, we can use the molar ratio between Fe and O2, which is 4:3, to calculate the number of moles of Fe

4 Fe = 3 O2

x Fe = 0.018 O2

x = 0.024 moles of Fe

Now we use the number of moles, and molar mass of Fe, 55.84g/mol to find the final mass of Fe:

55.84g = 1 mol

x grams = 0.024 moles

x = 1.34 grams of Fe is required