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?

Unbalanced equation: Na + HF “yields”/ NaF + H2

1.66 mol H2

2.49 mol H2

2.88 mol H2

3.33 mol H2

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?

Unbalanced equation: Na + HF “yields”/ NaF + H2

1.66 mol H2

2.49 mol H2

2.88 mol H2

3.33 mol H2

wolfe8 · Answer

First, balance the equation. Then, can you find the amount of H2 theoretically?

anonymous · Answer

Balanced equation: 2Na +2HF → 2NaF + H
molar ratio of Na: H2 is 2:1
5.75-mol Na will theoretically produces 5.75-mol/2 = 2.875-mol H2. 
The yield is 86.5%, so 2.875-mol x .865 = 2.49-mol H2.
Is that right?

wolfe8 · Answer

Looks right to me

anonymous · Answer

So the answer would be b?

wolfe8 · Answer

Yes. Tell me if it's wrong though.

anonymous · Answer

It was right

wolfe8 · Answer

Great!

wolfe8 · Answer

You would want to close this question.