Question

What volume of oxygen gas at 20°C and 0.919 atm is

consumed in the production of 5.00 g of iron(III) oxide from
metallic iron?
4Fe(s) + 302(g) → 2Fe2O3 (9)
LO2​

Answer 1

Volume of oxygen gas is 78.47L

Step-by-step explanation:

Data;

Temperature (T) = 20°C = (20 + 273.15)K = 293.15K

Pressure (P) = 0.919atm

Mass (m) = 5g

Volume (v) = ?

Equation of reaction;

4Fe + 3O₂ → 2Fe₂O₃

From the equation of reaction, the number of moles of oxygen is 3.

To find the volume of oxygen produced, we'll have to use ideal gas equation

PV = nRT

P = pressure

V = volume

n = number of moles

R = ideal gas constant = 0.082J/mol.K

T = temperature of the ideal gas

PV = nRT

V = nRT/ P

V = (3 × 0.082 × 293.15) / 0.919

V = 72.1149 / 0.919

V = 78.47L

Volume of the gas is 78.47L