Question

One mole of hydrogen gas is reacted with oxygen gas to produce water in this reaction:

2 H2 + O2 → 2 H2O

In the lab you actually made 14 grams of water. What is the percent yield?

35.60%

100%

11.11%

77.60%

Answer 1

77.6%

Step-by-step explanation:

Parameters given:

Number of moles of H₂ = 1 mole

Actual yield = 14g

Equation of the reaction:

2H₂ + O₂ → 2H₂O

Solution

From the statement of the problem, we have been given the actual yield that was obtained during the laboratory procedure. Now, we find the theoretical yield and this would lead us to the percentage yield.

Percentage yield =
`(actual yield)/(Theoretical yield)` x 100

Theoretical yield:

From the stoichiometeric equation:

2moles of H₂ produced 2 moles of water

Hydrogen gas the limiting reactant and it is in short supply,

Therefore, 1 mole of water would be produced by 1 mole of hydrogen gas.

We use this to estimate the mass of the water that would be produced:

mass of water = number of moles of water x molar mass of water

Molar mass of H₂O = (2x1) + 16 = 18gmol⁻¹

mass of water = 1 mol x 18gmol⁻¹ = 18g

This is the theoretical yield

Percentage yield =
`(actual yield)/(Theoretical yield)` x 100

Percentage yield =
`(14)/(18)` x 100 = 77.6%