Question

What volume of oxygen gas at 20 degrees C and 0.919 atm is consumed in the production of 5.00g of iron(III) oxide from metallic iron?? In liters

Answer 1

The volume of Oxygen gas at 20°C and 0.919 atm is 0.81 * 10^-5 L

Step-by-step 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