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For the reaction so3 + h2o h2so4, calculate the percent yield if 500. g of sulfur trioxide react with excess water to produce 575 g of sulfuric acid.

Respuesta :

kenmyna
The %  yield  if  500 g of  sulfur trioxide  reacted  with  excess  water to   produce  575 g  of  sulfuric  acid is calculated using  the  below  formula


%  yield = actual  yield/ theoretical  yield  x100

actual  yield =575 grams
to  calculate  theoretical  yield
find the  moles  of SO3   used =mass/molar  mass
=  500g/   80 g/mol =6.25  moles

SO3+H2O=H2SO4
by   use of  mole ratio  of SO3  :  H2SO4 which  is 1:1  the moles of H2SO4  is  also=  6.25  moles

the theoretical  yield of H2SO4 is therefore =  moles /molar  mass
=  6.25  x98=  612.5 grams

%yield  is therefore= 575 g/612 g   x100=  93.9  %

Eduard22sly

The percentage yield of the reaction is 93.9%

We'll begin by calculating the mass of SO₃ that reacted and the mass of H₂SO₄ produced from the balanced equation.

SO₃ + H₂O —> H₂SO₄

Molar mass of SO₃ = 32 + (3×16) = 80 g/mol

Mass of SO₃ from the balanced equation = 1 × 80 = 80 g

Molar mass of H₂SO₄ = (2×1) + 32 + (16×4) = 98 g/mol

Mass of H₂SO₄ from the balanced equation = 1 × 98 = 98 g

SUMMARY:

From the balanced equation above,

80 g of SO₃ reacted to produce 98 g of H₂SO₄

Next, we shall determine the theoretical yield of H₂SO₄

From the balanced equation above,

80 g of SO₃ reacted to produce 98 g of H₂SO₄.

Therefore,

500 g of SO₃ will react to produce = (500 × 98) / 80 = 612.5 g of H₂SO₄.

Finally, we shall determine the percentage yield.

  • Actual yield = 575 g
  • Theoretical yield = 612.5 g
  • Percentage yield =?

Percentage yield = (Actual /Theoretical) × 100

Percentage yield = (575 / 612.5) × 100

Percentage yield = 93.9%

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