Question

What volume, in liters, of O2 measured at 760 torr and 25o C is needed to completely react with 18.47 g of iron to give Fe2O3?

Answer 1

First calculate no of moles of Oxygen

Molar mass of Fe ₂ O ₃ = ( 2 x 55.85 ) +( 3 x 16) = 159.7 gm / mol

2 mole Fe 1 mole  Fe ₂ O ₃  

111.7 gm Fe   159.7 gm Fe ₂ O ₃  

18.47 gm Fe 159.7 gm x 18.47 gm / 111.7 gm Fe ₂ O ₃  

    26.41 gm Fe ₂ O ₃  

Again 1 mole  Fe ₂ O ₃ 3/2 mol O ₂

159.7 gm Fe ₂ O ₃   48 gm O ₂   

26.41 gm Fe ₂ O ₃   48 gm x 26.41 gm / 159.7 gm O ₂

   7.94 gm O ₂

Moles of   O ₂ = mass /molar mass

= 7.94 gm / 32 gm / mol

= 0.248 mols

Calculation for volume

We have ideal gas relation P V = n R T

where P pressure of a gas

V volume of a gas

n no of moles of a gas

R is gas constant

T is a temp. of a gas

Here P = 760 torr = 1 atm   

V = ?

n = 0.248

T = 298 K

R= 0.082057 L atm mol^-1 K^-1

Putting above values in gas equation we get

  P V = n R T

    V = n R T / P

= 0.248 mol x  0.082057 L atm mol^-1 K^-1 x 298 K / 1 atm

= 6.064 L