Question

• Consider the following reaction:

2 SO2(g) + O2(g) → 2 SO3(g)

If 285.5 mL of SO2 reacts with 158.9 mL of O2 (both measured at 0.06579 atm and 315K), what is the limiting reactant and what is the theoretical yieldof SO3?

Answer 1

Answer:

Given reaction is

2SO2(g) + O2(g) --------> 2SO3(g)

Given pressure, P=0.06579 atm, temperature, T=315 K

Volume of SO2, V=285.5 mL=0.2855 L and volume of O2, V=158.9 mL=0.1589 L

(Since 1 L=1000 mL)

From Ideal gas equation PV=nRT

Where n=number of moles and R=gas constant=0.08206 L atm mol^-1 K^-1.

Therefore moles of SO2=PV/RT=(0.06579 atm x 0.2955 L)/(0.08206 L atm mol^-1 K^-1 x 315 K)

Moles of SO2=7.266x10^-4 mol.

Moles of O2=(0.06579 mol x 0.1589 L)/(0.08206 L atm mol^-1 K^-1 x 315 K)

Moles of O2=4.04x10^-4 mol.

From balanced equation, mole ratio between SO2:O2=2:1

But here SO2:O2=(7.266x10^-4):(4.04x10^-4)=1.798:1

Therefore here SO2 is less, so it is limiting reactant.

Moles of SO3 formed=moles of SO2 used

(since SO2:SO3=2:2=1:1)

Therefore moles of SO3 formed=7.266 x10^-4 mol.

Molar mass of SO3=80.066 g/mol

Therefore mass of SO3=moles x molar mass=7.266 x10^-4 mol x 80.066 g/mol=0.0582 g.

Therefore theoretical yield SO3=0.0582 g.

Please let me know if you have any doubt. Thanks