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Oseni

When a reaction vessel initially containing 68.0 kg [tex]CH_4[/tex] and excess steam yields 16.9 kg [tex]H_2[/tex], the percent yield would be 49.8%

Percent yield

Recall that: percent yield = yield/theoretical yield x 100%

From the equation of the reaction:

[tex]CH_4 + 2H_2O --- > CO_2 + 4H_2[/tex]

The mole ratio of methane to steam is 1:4.

Mole of 68.0 kg methane = 68000/16.0 = 44239.4 moles

Equivalent mole of  [tex]H_2[/tex]= 16,957.6 moles

Mass of 16,957.6 moles  [tex]H_2[/tex] = 16,957.6 x 2 = 33,915.2 grams or 33.92 kg

Percent yield = 16.9/33.92 x 100% = 49.8%

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Eduard22sly

The percentage yield of the reaction that yields 16.9 kilograms of Hydrogen gas, H₂ is 66.3%

Balanced equation

CH₄ + H₂O —> CO + 3H₂

Molar mass of CH₄ = 16 g/mol

Mass of CH₄ from the balanced equation = 1 × 16 = 16 g = 0.016 Kg

Molar mass of H₂ = 2 g/mol

Mass of H₂ from the balanced equation = 3 × 2 = 6 g = 0.006 Kg

SUMMARY

From the balanced equation above,

0.016 Kg of CH₄ reacted to produce 0.006 Kg of H₂

How to determine the theoretical yield of H₂

From the balanced equation above,

0.016 Kg of CH₄ reacted to produce 0.006 Kg of H₂

Therefore,

68 Kg of CH₄ will react to produce = (68 × 0.006) / 0.016 = 25.5 Kg of H₂

How to determine the percentage yield

  • Actual yield of H₂ = 16.9 Kg
  • Theoretical yield of H₂ = 25.5 Kg
  • Percentage yield =?

Percentage yield = (Actual / Theoretical) × 100

Percentage yield = (16.9 / 25.5) ×100

Percentage yield = 66.3%

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