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

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

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 = $\frac{1}{2}\times 5.75=2.875mol$ of hydrogen gas.

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

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$

% yield of reaction = 86.5 %

Theoretical yield of hydrogen gas = 2.875 moles

Putting values in above equation, we get:

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

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.

Answer 2

Answer:

2.49mol H2 you're welcome

Explanation: