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Stoichiometry and percent yield of a zinc reaction

SCH3U Grade 11 Chemistry · Quantities in Chemical Reactions

The bench is laid out for a single-replacement run: an Erlenmeyer flask of dilute hydrochloric acid under a one-holed stopper, a delivery tube feeding a clamped gas syringe, and a crystallising dish waiting on the hot plate for the second half of the experiment. A student polishes a strip of zinc with emery paper, weighs it at 5.00 g, and drops it into the excess acid: Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g). The acid fizzes at once, the syringe plunger creeps steadily outward as the hydrogen comes off, and after some minutes the last grey sliver of metal dissolves and the bubbling dies away. The clear solution is poured into the dish and evaporated to dryness, and the white crust left behind weighs in at 9.61 g of zinc chloride. Taking the molar masses as 65.38 g/mol for zinc and 136.29 g/mol for zinc chloride, find the moles of zinc reacted, the volume of hydrogen released at STP, the mass of zinc chloride the equation predicts, and the percent yield the student achieved.

5.00 g Znexcess HCl(aq)H₂ 1.71 L @ STP9.61 g ZnCl₂ · 92.2% yield

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given
  • m = 5 g — Strip of zinc
  • m_act = 9.61 g — Zinc chloride recovered on evaporation
  • M_Zn = 65.38 g/mol — Molar mass of zinc
  • M_ZnCl₂ = 136.29 g/mol — Molar mass of zinc chloride
Determine
  1. (a)the moles of zinc reacted
  2. (b)the volume of hydrogen released at STP
  3. (c)the mass of zinc chloride the equation predicts
  4. (d)the percent yield achieved
Step 1 of 4(a) · solve for Amount of substance

Grams are what the balance reads; moles are what the balanced equation counts. Nothing else in the chain can start until the 5.00 g becomes an amount of substance.

mMn
Rearranged for n
n=mMn = \frac{m}{M}
Your values, in your units
n=(5 g)(65.38 g/mol)n = \frac{\left( 5\ \text{g} \right)}{\left( 65.38\ \text{g/mol} \right)}
Converted to base units
n=(0.005 kg)(65.38 g/mol)n = \frac{\left( 0.005\ \text{kg} \right)}{\left( 65.38\ \text{g/mol} \right)}
Answer
n=76.476 mmoln = 76.476\ \text{mmol}

Carried onward at full precision, not this rounded figure.

Open the Moles from Mass (n = m/M) solver →

Step 2 of 4(b) · solve for Gas volume at STP

Zinc and hydrogen stand one to one in the equation, so those same moles are moles of H₂ — about 1.71 L at STP. Nothing later needs this, but it is the quantity the gas syringe actually measured.

Vn
Rearranged for V
V=n VmV = n \, V_m
76.476 mmolcarried from step 1
Your values, in your units
V=(0.076476 mol) VmV = \left( 0.076476\ \text{mol} \right) \, V_m
Answer
V=1.7141 LV = 1.7141\ \text{L}

Carried onward at full precision, not this rounded figure.

Open the Gas Volume at STP solver →

Step 3 of 4(c) · solve for Mass

Zinc to zinc chloride is also one to one, so the moles carry straight across and only the molar mass changes. This is the theoretical yield — the most the flask could ever have given.

mMn
Rearranged for m
m=n Mm = n \, M
76.476 mmolcarried from step 1
Your values, in your units
m=(0.076476 mol) (136.29 g/mol)m = \left( 0.076476\ \text{mol} \right) \, \left( 136.29\ \text{g/mol} \right)
Answer
m=10.423 gm = 10.423\ \text{g}

Carried onward at full precision, not this rounded figure.

Open the Moles from Mass (n = m/M) solver →

Step 4 of 4(d) · solve for Percent yield

Now set the 9.61 g actually weighed beside the prediction. Both masses must be of the same substance — the product, never the reactant.

mtheoreticalmactual% yield
Rearranged for % yield
% yield=mactualmtheoretical×100%\%\,\text{yield} = \frac{m_{\text{actual}}}{m_{\text{theoretical}}} \times 100\%
10.423 gcarried from step 3
Your values, in your units
% yield=(9.61 g)(0.0104229 kg)×100%\%\,\text{yield} = \frac{\left( 9.61\ \text{g} \right)}{\left( 0.0104229\ \text{kg} \right)} \times 100\%
Converted to base units
% yield=(0.00961 kg)(0.0104229 kg)×100%\%\,\text{yield} = \frac{\left( 0.00961\ \text{kg} \right)}{\left( 0.0104229\ \text{kg} \right)} \times 100\%
Answer
% yield=92.201 %\%\,\text{yield} = 92.201\ \text{\%}

Carried onward at full precision, not this rounded figure.

Open the Percent Yield solver →

Answer

Therefore the strip held 0.0765 mol of zinc, the reaction released about 1.71 L of hydrogen at STP, the equation predicts 10.42 g of zinc chloride — and the 9.61 g on the balance is a 92.2% yield.

Why this order

Every stoichiometry problem is the same three-legged journey: grams in, moles across, grams out. The reason it has to run in that order is that a balanced equation says nothing whatsoever about mass — it counts particles. Students who try to shortcut from 5.00 g of zinc to grams of product by a mass ratio get an answer that is wrong by exactly the ratio of the molar masses, and the error is invisible because it still looks like a mass.

The one-to-one ratios here are doing quiet work. Zn : H₂ and Zn : ZnCl₂ are both 1 : 1, so the moles from step 1 feed steps 2 and 3 untouched. Swap in aluminium — 2Al + 6HCl → 2AlCl₃ + 3H₂ — and the hydrogen step needs a factor of 3/2 inserted by hand between the steps, because no solver on this site knows your balanced equation. The other trap is step 4: a percent yield above 100% is not a triumph but a diagnosis, almost always a product that was still damp when it was weighed. Zinc chloride is aggressively hygroscopic, so this particular experiment punishes a short drying time harder than most.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.