Question

If the reaction of 5.75 moles of sodium with excess hydrofluoric acid produced an 86.5% yield of hydrogen gas, what was the actual yield of hydrogen gas?

Answer 1

2.49mol H2 you're welcome

Step-by-step explanation:

Answer 2

Answer: The actual yield of hydrogen gas in the following reaction is 4.974 grams.

Step-by-step explanation:

The chemical equation for the reaction of sodium and hydrofluoric acid follows:

`2Na+2HF\rightarrow 2NaF+H_2`

As, hydrofluoric acid is present in excess. So, it is considered as an excess reagent and sodium is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

2 moles of sodium is producing 1 mole of hydrogen gas.

So, 5.75 moles of sodium will produce =
`(1)/(2)* 5.75=2.875mol` of hydrogen gas.

To calculate the actual yield of hydrogen gas, we use the equation:

`\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}* 100`

% yield of reaction = 86.5 %

Theoretical yield of hydrogen gas = 2.875 moles

Putting values in above equation, we get:

`86.5=\frac{\text{Actual yield of hydrogen gas}}{2.875moles}* 100\\\\\text{Actual yield of hydrogen gas}=2.487mol`

Now, to calculate the mass of hydrogen gas, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

Moles of hydrogen gas = 2.487 mol

Molar mass of hydrogen gas = 2g

Putting values in above equation, we get:

`2.487mol=\frac{\text{Mass of hydrogen gas}}{2g/mol}\\\\\text{Mass of hydrogen gas}=4.974g`

Hence, the actual yield of hydrogen gas in the following reaction is 4.974 grams.