I placed 1 gram of NaBH4 in a balloon and placed it over a glass round bottle filled with water and acetic acid. While secured, I emptied the balloon into the bottle and made sure to wash the inside of the balloon. The reaction taking place is $\ce{NaBH4 + 2(H2O) -> NaBO2 + 4(H2)}$.
The yield in this case should be 213 grams of $\ce{H2}$. According to Wikipedia, "hydrogen" has a density of 0.08988 g/L. That should leave me with 2.369 L of $\ce{H2}$. But my little experiment showed a balloon with the about 13 cm in diameter, which a volume of sphere that wide would be 1.15 liters of H2. This would mean a yield of .103 grams or 48.5%. But that just doesn't make sense. My NaBH4 has not degraded by half.
If you were to double the given density of $\ce{H2}$ to 0.17976 g/L, the results are much more in line with what I'd expect. The yield of $\ce{H2}$ would be .206 grams or 97%, which would make sense since not everything is ideal and my $\ce{NaBH4}$ is a bit old. Obviously there will be minor errors in measurement, and its not a perfect sphere and I didn't control for temperature, but that cannot account for a 50% reduction in yield. Maybe the balloon is compressing the gas? But I don't think so.
So the only other thing I can think of is that the presented density of 0.08988 g/L for hydrogen is technically only for monoatomic Hydrogen, or $\ce{H}$, which doesn't make sense and no one ever uses. The density of $\ce{H2}$ at STP should be 0.17976 g/L then to match my observations. There must be something I'm missing here entirely that I'm not understanding.