Answer:
The volume of Oxygen gas at 20°C and 0.919 atm is 0.81 * 10^-5 L
Explanation:
P V = nRT
n = mass / molar mass
P V = mRT / Mm
V = mRT / P Mm
m = 5 g
Mm of Fe2O3 = ( Fe = 55.8, O = 16) = ( 55.8 * 2 + 16 * 3) = 111.6 + 48 = 159 6 g/mol
P = 0.919 atm = 0.919 * 1.01 10 ^5 Nm^-2 = 0.928 * 10^5 Nm^-2
R = 0.082 L atm mol^-1 K^-1
T = 20°C = 20 + 273 K = 293 K
So therefore,
V = 5 * 0.082 * 293 / 0.928 * 10^5 * 159.6
V = 120.13 / 148.1088 * 10^5
V = 0.81 * 10^-5 L
The volume of oxygen gas consumed is 0.81 * 10^-5 L
