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.
Good practice is starting with symbolic algebraic expressions and keeping it this way until all is ready to plug in literal numbers. It helps in focusing on principles, mistakes are easier to spot, orientation is improved, Q/A is reusable and has bigger permanent value.
You may find useful formatting mathematical/chemical expressions/formulas.
– Poutnik Dec 30 '22 at 07:07