Hydrogen Collected over Water: Limiting Reagent and Vapour-Pressure Corrections
A student is asked to prepare hydrogen and measure the yield. She weighs out 0.0500 mol of zinc granules, adds 25.0 mL of 2.00 mol L⁻¹ HCl, and collects the gas over water in an inverted, graduated burette at 22.0 °C. The room barometer reads 756 mmHg; the vapour pressure of water at 22.0 °C is 19.8 mmHg. She levels the burette before reading and records a gas volume of 0.630 L (R = 0.08206 L atm mol⁻¹ K⁻¹).
She expects 0.0500 mol of H2 - 'one mole of zinc gives one mole of hydrogen' - and is surprised that her calculation from the collected volume gives a much smaller number. She suspects the zinc was impure and wants to report a poor yield. In a second run at 30.0 °C, where the vapour pressure of water is 31.8 mmHg, she plans to use the same amounts and to compare the two gas volumes directly to see whether 'more gas' is produced when it is warmer.
1. Calculate the amount of hydrogen from the corrected data, identify the limiting reagent, and reconcile the result with the student's expectation.
Read the worked answer
Correct the pressure first. The gas in the burette is saturated with water vapour, so the hydrogen partial pressure is
p(H2) = 756 − 19.8 = 736.2 mmHg = 736.2/760 = 0.9687 atm
Temperature: T = 22.0 + 273.15 = 295.15 K.
Amount of gas: n = pV/RT = (0.9687 × 0.630) / (0.08206 × 295.15) = 0.6103 / 24.22 = 0.0252 mol of H2.
Limiting reagent: Zn + 2HCl → ZnCl2 + H2. n(HCl) = 0.0250 L × 2.00 mol L⁻¹ = 0.0500 mol, which can give at most 0.0500/2 = 0.0250 mol H2. Zinc is present as 0.0500 mol and would need 0.100 mol HCl, so HCl is limiting and zinc is in excess (about twice the stoichiometric amount).
Reconciliation: the measured 0.0252 mol agrees with the 0.0250 mol prediction to within 1%, well inside experimental error. The student's error was stoichiometric, not experimental: she forgot the 2:1 HCl:H2 ratio and assumed zinc limited the reaction. No impurity needs to be invoked, and the 'poor yield' is an artefact of dividing by the wrong theoretical value.
Note that if she had forgotten the water-vapour correction she would have obtained n = (756/760)(0.630)/24.22 = 0.0259 mol, about 3.5% too high, and would have reported a yield slightly above 100%.
2. Is zinc purity a reasonable explanation for the shortfall? Which remaining experimental errors would bias the result high, and which low?
Read the worked answer
No - there is no shortfall to explain. The corrected result (0.0252 mol) is slightly above the theoretical 0.0250 mol, so an impurity in the zinc cannot be the cause; impurities such as zinc oxide or lead would reduce the hydrogen produced, not raise it.
Errors that bias high (reported n too large):
- Omitting the water-vapour correction, as above: +3.5% at 22 °C. This is the largest single systematic error and is always in one direction.
- Reading the burette before levelling the inner and outer liquid levels; if the inner level is lower, the pressure inside is less than atmospheric.
- A wet, cold or un-equilibrated gas: if the gas is still cooler than 22 °C when read, n = pV/RT is overestimated.
- Loss of HCl by evaporation, making the real acid concentration lower - that would lower, not raise, the result, so it does not explain the excess either.
Errors that bias low:
- Hydrogen dissolving in the water and in the acid (small but real, ~0.02% solubility, negligible here).
- A leaky bung or a gas bubble escaping during collection.
- Zinc granules not fully consumed because of oxide coating - irrelevant here, since HCl limits.
The honest judgement is: with a 1% discrepancy between 0.0250 and 0.0252 mol, no systematic error needs to be postulated at all. The student should report the yield against the limiting reagent and note the vapour-pressure correction explicitly.
3. Predict the collected gas volume in the 30.0 °C run with the same amounts of reagents, and evaluate the student's plan to compare the two volumes directly.
Read the worked answer
Recompute with the new temperature and new vapour pressure. The amount of hydrogen is still set by the limiting HCl: n = 0.0250 mol.
p(H2) = 756 − 31.8 = 724.2 mmHg = 724.2/760 = 0.9529 atm T = 30.0 + 273.15 = 303.15 K
V = nRT/p = (0.0250 × 0.08206 × 303.15) / 0.9529 = 0.62195 / 0.9529 = 0.653 L
So the volume rises from 0.630 L to about 0.653 L, a 3.6% increase - even though the same 0.0250 mol of gas was produced.
Why the student's comparison is invalid. The volume of a gas is not a measure of how much gas there is: V depends on T and p as well as on n. Both runs produced identical amounts, yet the warmer run shows a bigger volume. Two corrections are needed before any comparison:
- Temperature, via the ideal-gas law - the dominant but not the only effect here.
- Water vapour, whose contribution grows faster than the gas volume (19.8 → 31.8 mmHg), which slightly offsets the temperature effect.
Only by converting both readings to moles (n = p_dry·V/RT, with p_dry = p_bar − p_water) is the comparison meaningful. This is the general lesson for collecting gases over water: the wet volume is a property of the surroundings, not of the reaction, and the vapour pressure of water must always be subtracted before reaching any conclusion about yield, molar mass or stoichiometry.