2. If the reaction of 5.75 moles of sodium with excess hydrofluoric acid is able to produce 2.49 mol H2, what is the percent yield of hydrogen gas? Unbalanced equation:
Na + HF –> NaF + H2

94.2%
86.5%
43.3%
23.1%

86.5%

Explanation:

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

2Na + 2HF –> 2NaF + H2

Next, we shall determine the theoretical yield of the hydrogen gas, H2. This is illustrated:

From the balanced equation above,

2 moles of Na reacted to produce 1 mole of H2.

Therefore, 5.75 moles of Na will react to produce = (5.75 x 1)/2 = 2.88 moles of H2.

Therefore, the theoretical yield of Hydrogen gas, H2 is 2.88 moles.

Finally, we shall determine the percentage yield of Hydrogen gas, H2. This can be obtained as follow:

Actual yield of H2 = 2.49 moes

Theoretical yield of H2 = 2.88 moles

Percentage yield of H2 =.?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 2.49/2.88 x 100

Percentage yield = 86.5%

Therefore, the percentage yield of Hydrogen gas, H2 is 86.5%.

the answer for sodium is 0.186

the answer for rubidium is 0.251