Worked examples — Preparation, properties, uses of dihydrogen

This page is a problem gym. The parent note taught you the reactions and the three big laws (stoichiometry, Faraday's law, the ideal-gas volume $22.4\text{L/mol}$). Here we drill them until no exam wording can surprise you.

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The scenario matrix

Before the notation, three plain-word tools we will lean on every time:

Think of a problem as spinning gears: grams → moles → (mole-ratio) → moles → grams or litres.

Here is the full space this topic can throw at you:

Cell	Scenario class	The twist it tests
A	Metal + acid, simple 1:1	basic grams→litres chain
B	Mole-ratio ≠ 1:1 (Al + acid, 2:3)	balancing forces a fraction
C	Limiting reagent (true excess case)	which reactant runs out first
D	Zero / degenerate input	conc. acid or excess → answer is not what you expect
E	Industrial steam reforming (1:4) + efficiency	theoretical vs actual yield
F	Electrolysis, Faraday's law	charge $Q=It$, $z=2$ electrons
G	Real-world word problem (fuel-cell car)	reading, unit juggling
H	Exam twist (Graham's law + non-STP gas volume)	property applied, not preparation

Eight worked examples below — one per cell.

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Cell A — Metal + acid, 1:1 ratio

$\text{Zn}+2\text{HCl}\to {\text{ZnCl}}_2+{\text{H}}_2$

1. Moles of Zn $=\frac{3.25}{65}=0.05\text{mol}$. Why this step? grams → moles is the only way to enter the mole-ratio gear ($M_{\text{Zn}}=65$).
2. Mole ratio Zn : H₂ = 1 : 1, so $n_{{\text{H}}_2}=0.05\text{mol}$. Why this step? the balanced equation fixes the exchange rate between reactant and product.
3. Volume $=0.05\times 22.4=1.12\text{L}$. Why this step? moles → litres via molar volume ($22.4\text{L/mol}$), which is only valid because the problem specifies STP.

Verify: $0.05\text{mol}$ is one-twentieth of a mole, and $22.4/20=1.12$. ✓ Units: mol × (L/mol) = L. ✓ Under a litre as forecast.

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Cell B — Mole ratio is not 1:1

$2\text{Al}+3{\text{H}}_2{\text{SO}}_4\to {\text{Al}}_2({\text{SO}}_4{)}_3+3{\text{H}}_2$

1. Moles of Al $=\frac{5.4}{27}=0.2\text{mol}$. Why? enter the mole gear ($M_{\text{Al}}=27$).
2. Ratio Al : H₂ = 2 : 3, so $n_{{\text{H}}_2}=0.2\times \frac{3}{2}=0.3\text{mol}$. Why? here the coefficients are unequal — you must carry the fraction $\frac{3}{2}$, not assume 1:1. This is the whole point of Cell B.
3. Volume $=0.3\times 22.4=6.72\text{L}$.

Verify: $0.2\text{mol}$ Al should give $1.5\times$ its own moles of H₂ = $0.3$. And $0.3\times 22.4=6.72\text{L}$. ✓ More gas per gram than zinc, as forecast (Al is light and trivalent).

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Cell C — Limiting reagent (true excess case)

$\text{Zn}+2\text{HCl}\to {\text{ZnCl}}_2+{\text{H}}_2$

1. Moles of each reactant. $n_{\text{Zn}}=\frac{6.5}{65}=0.10\text{mol}$; $n_{\text{HCl}}=1\text{M}\times 0.100\text{L}=0.10\text{mol}$. Why? you cannot compare who runs out without both in moles.
2. Divide each by its coefficient. The equation needs 2 HCl per 1 Zn. $\frac{0.10}{1}=0.10$ (Zn) vs $\frac{0.10}{2}=0.05$ (HCl). The smaller number wins: HCl is the limiting reagent, and $0.05\text{mol}$ of Zn is left over unreacted. Why? dividing by the coefficient converts "moles I have" into "reaction batches I can run"; whoever supports fewer batches runs out first.
3. Product from the limiting reagent. HCl : H₂ = 2 : 1, so $n_{{\text{H}}_2}=\frac{0.10}{2}=0.05\text{mol}$; $V=0.05\times 22.4=1.12\text{L}$. Why? the product is dictated by whoever runs out — the leftover Zn produces nothing extra.

Verify: using the excess reactant (Zn) would wrongly predict $0.10{\text{mol H}}_2$ = $2.24\text{L}$ — but there isn't enough acid, so the true answer is the smaller $1.12\text{L}$. ✓ See Redox Reactions for why Zn donates exactly 2 electrons.

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Cell D — Degenerate input (the trap)

1. Recognise the regime change. Concentrated ${\text{H}}_2{\text{SO}}_4$ is an oxidising agent — its sulfur, not its H⁺, gets reduced: $\text{Zn}+2{\text{H}}_2{\text{SO}}_4(\text{conc})\to {\text{ZnSO}}_4+{\text{SO}}_2+2{\text{H}}_2\text{O}$ Why this step? the "which reaction?" question comes before any arithmetic. A degenerate input can change the reaction itself.
2. Count the H₂. There is no H₂ on the product side — the hydrogen leaves as water. $n_{{\text{H}}_2}=0\text{mol}\Rightarrow V=0\text{L}.$

Verify: the whole point of "use dilute acid" (parent note [!mistake]) is exactly this: the degenerate answer is zero. ✓ Always ask which reaction runs before you compute.

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Cell E — Industrial: 1:4 ratio + efficiency

${\text{CH}}_4+2{\text{H}}_2\text{O}\to {\text{CO}}_2+4{\text{H}}_2$

1. Moles of CH₄ $=\frac{80000\text{g}}{16\text{g/mol}}=5000\text{mol}$. Why? kilograms → grams → moles; work in grams to keep $M$ tidy.
2. Ratio 1:4 → theoretical $n_{{\text{H}}_2}=5000\times 4=20000\text{mol}$. Why? the net equation (from adding the two steps, see downstream user of this H₂) fixes 4 H₂ per CH₄.
3. Theoretical mass $=20000\times 2=40000\text{g}=40\text{kg}$. Why? moles → grams by multiplying by molar mass ($M_{{\text{H}}_2}=2\text{g/mol}$); this is the reverse of the grams→moles gear used in step 1.
4. Apply efficiency $=40\times 0.75=30\text{kg}$. Why? efficiency multiplies the ideal mass; it never changes the stoichiometry.

Verify: $80\text{kg}$ CH₄ $\to$ ideal $40\text{kg}$ H₂ (a clean 1:2 mass ratio because $16\times 1$ vs $2\times 4=8$… check: $8/16=0.5$, so ideal H₂ = half the CH₄ mass = $40\text{kg}$ ✓). Then $30\text{kg}$ after loss. ✓

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Cell F — Electrolysis via Faraday's law

Figure F — a blueprint of the electrolysis cell. Amber arrows = the two electrons $2e^{-}$ arriving at the cyan cathode on the left; the white arrow = two ${\text{H}}^{+}$ ions drifting in; the cyan circle rising off the electrode = one ${\text{H}}_2$ bubble leaving. The picture is the reason $z=2$ sits in the denominator of Faraday's law: it visibly takes two electrons to assemble one molecule.

Look at the figure: it takes two electrons to build one ${\text{H}}_2$ molecule — that factor $z=2$ is why it sits in the denominator below.

$n=\frac{Q}{zF}=\frac{It}{zF},F=96485\text{C/mol}$

1. Charge $Q=It=10\times (30\times 60)=18000\text{C}$. Why? seconds, not minutes — the ampere is coulombs per second.
2. Moles of H₂ $=\frac{18000}{2\times 96485}=0.0933\text{mol}$. Why? $z=2$: each H₂ swallows two electrons at the cathode (see Electrolysis and Faraday's Laws).
3. (a) Mass $=0.0933\times 2=0.187\text{g}$. Why? moles → grams by multiplying by molar mass ($M_{{\text{H}}_2}=2\text{g/mol}$): each mole weighs 2 g.
4. (b) Volume $=0.0933\times 22.4=2.09\text{L}$ at STP. Why? moles → litres via the molar volume $22.4\text{L/mol}$, valid because the answer is requested at STP.

Verify: $18000/(2\times 96485)=0.0933$ ✓. Sanity: about a tenth of a mole → about a tenth of $22.4$ = $2.2\text{L}$ ✓. This is the clean-H₂ route.

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Cell G — Real-world word problem

1. Moles of H₂ needed $=\frac{1000\text{g}}{2\text{g/mol}}=500\text{mol}$. Why? start from the target and work backwards through the gears.
2. Charge required $Q=nzF=500\times 2\times 96485=9.6485\times 10^7\text{C}$. Why? invert Faraday's law: to make $n$ mol you must pass $nzF$ coulombs.
3. Time $t=\frac{Q}{I}=\frac{9.6485\times 10^7}{200}=482425\text{s}$. Why? the definition $Q=It$ rearranges to $t=Q/I$; we know the charge and the current, so solving for time just divides.
4. Convert to hours $=482425/3600\approx 134\text{h}$. Why? the question asked for hours; always land in the requested unit.

Verify: $500\text{mol}\times 2e^{-}$/mol $\times 96485=9.65\times 10^7\text{C}$ ✓; divided by $200\text{A}$ then $3600$ gives $\approx 134\text{h}$ ✓. That's why hydrogen refuelling uses stored gas, not on-the-spot electrolysis.

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Cell H — Exam twist: Graham's law + a non-STP gas volume

Part (a):

1. Molar masses $M_{{\text{H}}_2}=2$, $M_{{\text{O}}_2}=32$. Why? the ratio only cares about mass, not the reaction that made the gas.
2. Plug in $\frac{r_{{\text{H}}_2}}{r_{{\text{O}}_2}}=\sqrt{\frac{32}{2}}=\sqrt{16}=4$. Why? the square root turns a $16\times$ mass gap into a $4\times$ speed gap.

Part (b): here $22.4\text{L/mol}$ is forbidden — that shortcut is only for STP. We must use the full ideal-gas law $PV=nRT$ with $R=0.0821{\text{L⋅atm⋅mol}}^{-1}{\text{K}}^{-1}$. 3. Rearrange $V=\frac{nRT}{P}=\frac{0.50\times 0.0821\times 300}{2}=6.16\text{L}$. Why? off STP, volume depends on the actual $T$ and $P$; the molar-volume gear no longer applies.

Verify: (a) $\sqrt{32/2}=\sqrt{16}=4$ exactly ✓ — so ${\text{H}}_2$ diffuses 4× faster than ${\text{O}}_2$ (see Graham's Law of Diffusion). (b) $PV=nRT$ gives $6.16\text{L}$; compare the STP value $0.50\times 22.4=11.2\text{L}$ — the higher pressure more than cancels the higher temperature, so it shrinks, as forecast. ✓

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Recall Self-test — name the cell and the recipe

Which conversion sequence for a metal-acid problem? ::: grams → moles (÷$M$) → mole-ratio → moles → ×22.4 for litres. What does STP stand for, and what number does it unlock? ::: Standard Temperature ($273\text{K}$) and Pressure ($1\text{atm}$); it unlocks the $22.4\text{L/mol}$ molar volume. In Faraday's law, why is $z=2$ for H₂? ::: two electrons are needed at the cathode to reduce $2{\text{H}}^{+}$ into one ${\text{H}}_2$ molecule. How do you find the limiting reagent? ::: divide each reactant's moles by its coefficient; the smallest result runs out first. Concentrated ${\text{H}}_2{\text{SO}}_4$ on zinc gives how much H₂? ::: zero — it oxidises, releasing ${\text{SO}}_2$ and water instead. ${\text{H}}_2$ vs ${\text{O}}_2$ diffusion speed ratio? ::: 4 : 1, from $\sqrt{32/2}$.

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Prerequisites drilled here: Redox Reactions · Electrolysis and Faraday's Laws · Thermodynamics of Chemical Reactions · Graham's Law of Diffusion · Hydrogen - Introduction and Isotopes. 🇮🇳 Isko Hinglish mein padho →