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In: Chemistry

Consider the following reaction: 2SO2(g)+O2(g)→2SO3(g) Part B What is the theoretical yield of SO3? nSO3 =...

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

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.

Expert Solution

Given reaction is 2 SO2(g) + O2(g) → 2 SO3(g)

From this reaction, it is clear that 2 moles of SO₃ is produced.

B) Therefore,

theoretical yield of SO₃ = 2 moles

C) We have to calculate the no of moles of SO3 by using ideal gas equation.

Given that Volume of SO3, V= 175.8 mL = 0.1758 L

Temp T = 327 K

Pressure P = 48.7 mmHg = 48.7/ 760 atm = 0.064 atm [ 1 atm = 760 mmHg ]

no of moles of SO3, n = ?

We know that Ideal gas equation is

PV = nRT where R = Ideal gas constant = 0.0821 L.atm/K/mol

n = PV/RT

= [0.064 atm x 0.1758 L]/ [0.0821 L.atm/K/mol x 327 K ]

= 0.00042 moles

This is the actual yield of SO3.

Therefore,

percent yield of SO3 = [actual yield of SO3 / theoretical yield of SO3] x 100

= [0.00042 moles / 2 moles] x 100

= 0.021 %

queen_honey_blossom answered 2 years ago

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