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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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    $\begingroup$ Looks good to me. I would have calculated 523.6/22400= 0.023375 mole H2 for simplicity, but you got it right. $\endgroup$ Nov 7 '20 at 15:06
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    $\begingroup$ editing tip: use \pu{} for physical units and \ce{} for chemical formulae/equations. $\endgroup$ Nov 7 '20 at 16:06
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    $\begingroup$ @Kedar do not apologize to anyone or feel guilty about posting a genuine question. You are trying to help your son and this should be enough to convince anyone interested in science. $\endgroup$
    – M. Farooq
    Nov 8 '20 at 3:36
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Kedar's calculation may be right. But it must be realized that the reaction will be rather slow. Even with a $1 M$ solution, the reaction is slow. The reaction of $1.5$ $\pu{g}$ metallic zinc with such a diluted $\ce{HCl}$ solution ($0.1 M$) will last more than one hour. Anyway, in order to inflate such a balloon a higher pressure than the atmospheric pressure must be applied. If not, the balloon will keep its original dimension. How to increase the pressure of $\ce{H2}$ in such a balloon for inflating it, if the hydrogen is produced by the reaction of zinc on $\ce{HCl}$ ?

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