Question

If 4.50 L of water vapor at 50.2 °C and 0.121 atm reacts with excess iron, how many grams of iron(III)oxide will be produced? 2 Fe(s) + 3H,0() Fe,0,(s) + 3H, (p) mass

Answer 1

By ideal gas equation

PV = nRT

given , volume (V ) = 4.50 L

pressure (P) = 0.121 atm

R = 0.082 L-atm/mol.K

temperature (T) = 50.2 + 273 K = 323.2 K

then moles of hydrogen gas will be ; n = (PV/RT)

= (0.121$\times$4.50)/0.082$\times$323.2)

= 0.0205

from the balanced reaction

mole ratio of Iron (III) oxide to Hydrogen is 1 : 3

therefore moles of Iron (III) oxide formed

= 0.0205/3

= 0.0068

mass of Iron (III) oxide will be produced

= 0.0068 $\times$ 160 (g/mol)

= 1.09 g