Question

Consider the reaction below. If 8.75 g of CH4 reacts with 18.95 g of O2 and 6.85 g of H2O is collected, what is the percent yield of H2O

Answer 1

The percentage yield of water, H₂O obtained from the reaction is 64.3%

We'll begin by writing the balanced equation for the reaction. This is given below:

CH₄ + 2O₂ —> CO₂ + 2H₂O

• Next, we shall determine the masses of CH₄ and O₂ that reacted and the mass of H₂O produced from the balanced equation. This can be obtained as follow:

Molar mass of CH₄ = 12 + (4×1) = 16 g/mol

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

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mass of O₂ from the balanced equation = 2 × 32 = 64 g

Molar mass of H₂O = (2×1) + 16 = 18 g/mol

Mass of H₂O from the balanced equation = 2 × 18 = 36 g

SUMMARY:

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂ to produce 36 g of H₂O

• Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂.

Therefore,

8.75 g of CH₄ will react with = (8.75 × 64)/16 = 35 g of O₂

From the calculation made above, we can see that a higher amount of O₂ (i.e 35 g) than what was given (i.e 18.95 g) is needed to react with 8.75 g of CH₄.

Therefore, O₂ is the limiting reactant and CH₄ is the excess reactant.

• Next, we shall determine the theoretical yield of H₂O.

From the balanced equation above,

64 g of O₂ reacted to produce 36 g of H₂O.

Therefore,

18.95 g of O₂ will react to produce = (18.95 × 36)/64 = 10.66 g of H₂O.

Thus, the theoretical yield of H₂O is 10.66 g

• Finally, we shall determine the percentage yield of H₂O. This can be obtained as follow:

Actual yield of H₂O = 6.85 g

Theoretical yield of H₂O = 10.66 g

Percentage yield of H₂O =?

$Percentage yield = \frac{Actual}{Theoretical} * 100\\\\= \frac{6.85}{10.66} * 100$

= 64.3%

Therefore, the percentage yield of H₂O is 64.3%

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