Question

Consider the following reaction: 2SO2(g)+O2(g)→2SO3(g)

Part A If 276.0 mL of SO2 is allowed to react with 161.4 mL of O2 (both measured at 327 K and 48.7 mmHg ), what is the limiting reactant?

Part B What is the theoretical yield of SO3? nSO3 =

Part C If 175.8 mL of SO3 is collected (measured at 327 K and 48.7 mmHg ), what is the percent yield for the reaction? Express your answer using four significant figures.

Answer 1

Part A

First calculate mole of gas

no. of mole of SO₂

use ideal gas equation to calculate mole

Ideal gas equation

PV = nRT             where, P = atm pressure= 48.7 mmHg = 0.064 atm,

V = volume in Liter = 276 ml = 0.276 L

n = number of mole = ?

R = 0.08205L atm mol^-1 K^-1 =Proportionality constant = gas constant,

T = Temperature in K = 327 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (0.064X 0.276)/(0.08205X 327) = 0.000658 mole

mole of SO₂ = 0.000658 mole

no. of mole of O₂

use ideal gas equation to calculate mole

Ideal gas equation

PV = nRT             where, P = atm pressure= 48.7 mmHg = 0.064 atm,

V = volume in Liter = 161.4 ml = 0.1614 L

n = number of mole = ?

R = 0.08205L atm mol^-1 K^-1 =Proportionality constant = gas constant,

T = Temperature in K = 327 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (0.064X 0.1614)/(0.08205X 327) = 0.000385 mole

mole of O₂ = 0.000385 mole

According to reaction 2 mole of SO₂ react with 1 mole of O₂ therefore to react with 0.000385 mole of O₂ required SO₂ = 0.000385 X 2 = 0.00077 mole but given SO₂ is only 0.000658 mole therefore SO₂ is limiting reactant.

SO₂ is limiting reactant.

Part B

According to reaction 2 mole of SO₂ produce 2 mole of SO₃ therefore 0.000658 mole of SO₂ produce 0.000658 mole of SO₃

theoretical yield of SO₃ = 0.000658 mole

Part C

Calculate mole of SO₃

no. of mole of SO₃

use ideal gas equation to calculate mole

Ideal gas equation

PV = nRT             where, P = atm pressure= 48.7 mmHg = 0.064 atm,

V = volume in Liter = 175.8 ml = 0.1758 L

n = number of mole = ?

R = 0.08205L atm mol^-1 K^-1 =Proportionality constant = gas constant,

T = Temperature in K = 327 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (0.064X 0.1758)/(0.08205X 327) = 0.000419 mole

mole of SO₃ = 0.000419 mole

percent yield = actual yield X 100/ theoretical yield

percent yield of SO₃ = 0.000419 X 100 / 0.000658 = 63.68 %

percent yield of SO₃ = 63.68 %