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?

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:

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 = of hydrogen gas.

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

% yield of reaction = 86.5 %

Theoretical yield of hydrogen gas = 2.875 moles

Putting values in above equation, we get:

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

Moles of hydrogen gas = 2.487 mol

Molar mass of hydrogen gas = 2g

Putting values in above equation, we get:

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

c. because step up transformer steps up the input voltage i.e increases the input voltage  

the number of turns are greater on the secondary side of step up transformer  

transformer does not handle dc current