I was a bit hesitant to post this question for it may appear like a "homework help" type of question. But upon rethinking it felt like a genuine question. I was studying stoichiometry with my son (from the OpenStax Chemistry 2e book) and suddenly I had an idea to use these calculations in a real chemistry experiment that I am going to perform at home (of course, I truly believe in a "safety first" approach). I wrote it all up here on GitHub, but reposting here to keep it all at one place:
A practical problem: It was decided to inflate a common latex balloon to 5cm radius by the hydrogen gas produced during a single-displacement reaction: $$\ce{2HCl(aq) + Zn(s) -> ZnCl2(s) + H2(g)}$$ that uses an N/10 i.e. 0.1M HCl solution (available in many drugstores, typically in $\pu{200ml}$ bottles). How much reactants are needed?
Solution: We need to inflate the balloon to about $\pu{5cm}$ radius. Assuming the balloon to be a sphere gives us the volume of balloon, $V_b$, as: $V_b = \frac{4}{3}\cdot \pi r^3 = \pu{523.599cm^3}$.
Assuming the density of hydrogen at STP as 0.0000899 $\pu{g/cm^3}$, the mass of hydrogen is $\pu{523.599 \times \space 0.0000899 g} = \pu{0.0471 g}$. Since 1 mol $\ce{H_2(g)}$ has a mass of $\pu{2 \times \space 1.008 g} = \pu{2.016 g}$, we need $\pu{\frac{0.0471}{2.016} mol} = \pu{0.0233 mol}$ $\ce{H_2(g)}$.
Balancing the reaction, we see that to produce 1mol $\ce{H_2(g)}$ we need 2 mol HCl(aq) and 1 mol Zn(s).
This means that to produce 0.0233 mol $\ce{H_2(g)}$ we need $2 \times 0.0233 = \pu{0.0466 mol}\ \ce{HCl(aq)}$. We have a 0.1M HCl (aqueous) solution which means that $\frac{\text{mol of solute}}{\text{litres of solution}} = 0.1$. Clearly, we need $\frac{0.0466}{0.1} = \pu{0.466 L}$ or $\pu{466 ml}$ HCl.
The amount of zinc needed = $0.0233 \times \text{molar mass of zinc} = 0.0233 \times 65.38 = \pu{1.523 g}$.
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for chemical formulae/equations. $\endgroup$