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General Chemistry covers stoichiometry, atomic structure, bonding, gases, thermochemistry, equilibrium, and acid-base chemistry. Exams are calculation-heavy with units and significant figures graded, and conceptual questions that punish memorization without understanding. Timed practice with worked solutions is the reliable way to get both halves exam-ready.

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Topics typically tested

  • Stoichiometry & limiting reagents
  • Atomic structure & periodicity
  • Chemical bonding & molecular geometry
  • Gas laws
  • Thermochemistry & Hess's law
  • Chemical equilibrium & acid-base chemistry

Sample General Chemistry exam questions

Three original questions in the style of a real General Chemistry exam, each with a worked solution. Your own practice exam can be built from your course's notes and syllabus.

Question 1Multiple choiceStoichiometry & limiting reagents

Aluminium reacts with chlorine gas according to

2 Al(s)+3 Cl2(g)→2 AlCl3(s)2\,\mathrm{Al}(s) + 3\,\mathrm{Cl_2}(g) \rightarrow 2\,\mathrm{AlCl_3}(s)

A sample of 10.0 g of aluminium is placed in a sealed vessel with 35.0 g of chlorine gas and the reaction goes to completion. What mass of aluminium chloride is produced? (Molar masses: Al 26.98 g/mol, Cl 35.45 g/mol.)

  1. A.

    43.9 g

  2. B.

    49.4 g

  3. C.

    65.8 g

  4. D.

    98.7 g

Worked solution
Answer:A

Convert both reactants to moles, because the limiting reagent is decided by mole ratio, not by mass.

nAl=10.0 g26.98 g/mol=0.3706 mol,nCl2=35.0 g70.90 g/mol=0.4937 moln_{\mathrm{Al}} = \frac{10.0\ \text{g}}{26.98\ \text{g/mol}} = 0.3706\ \text{mol}, \qquad n_{\mathrm{Cl_2}} = \frac{35.0\ \text{g}}{70.90\ \text{g/mol}} = 0.4937\ \text{mol}

Compare with the 2 : 3 stoichiometry. To consume all 0.3706 mol of Al would require 0.3706×32=0.556 mol0.3706\times\tfrac{3}{2} = 0.556\,\text{mol} of Cl2\mathrm{Cl_2}, but only 0.4937 mol is available, so chlorine is the limiting reagent even though it is the heavier sample.

Product moles follow from the limiting reagent and the 3 : 2 ratio between Cl2\mathrm{Cl_2} and AlCl3\mathrm{AlCl_3}:

nAlCl3=0.4937 mol Cl2×2 mol AlCl33 mol Cl2=0.3291 moln_{\mathrm{AlCl_3}} = 0.4937\ \text{mol}\ \mathrm{Cl_2} \times \frac{2\ \text{mol}\ \mathrm{AlCl_3}}{3\ \text{mol}\ \mathrm{Cl_2}} = 0.3291\ \text{mol}

Convert to mass with M(AlCl3)=26.98+3(35.45)=133.33 g/molM(\mathrm{AlCl_3}) = 26.98 + 3(35.45) = 133.33\ \text{g/mol}:

m=0.3291 mol×133.33 g/mol=43.9 gm = 0.3291\ \text{mol} \times 133.33\ \text{g/mol} = 43.9\ \text{g}

As a check, the aluminium left over is 0.3706−0.3291=0.0415 mol0.3706-0.3291 = 0.0415\,\text{mol} (1.12 g), a positive amount, which confirms that aluminium was in excess.

Mass of AlCl3\mathrm{AlCl_3} formed: 43.9 g (option A).

Option B (49.4 g) is what you get by treating aluminium as the limiting reagent because it has the smaller mass; the comparison must be made in moles using the coefficients. Option C (65.8 g) uses a 1 : 1 ratio between Cl2\mathrm{Cl_2} and AlCl3\mathrm{AlCl_3}, and option D (98.7 g) inverts the 2 : 3 ratio.

Question 2Short answerGas laws

What mass of oxygen gas, O2\mathrm{O_2}, occupies 5.00 L at 22.0 °C and 98.5 kPa? Treat oxygen as an ideal gas, use R=8.314 L kPa mol−1 K−1R = 8.314\,\mathrm{L\,kPa\,mol^{-1}\,K^{-1}}, and take the molar mass of O2\mathrm{O_2} as 32.00 g/mol.

Worked solution
Answer:

6.42 g

Use the ideal gas law PV=nRTPV = nRT to find the amount of gas, then convert moles to mass. The temperature must be in kelvin, because the gas law is a proportionality measured from absolute zero:

T=22.0+273.15=295.15 KT = 22.0 + 273.15 = 295.15\ \text{K}

Solve for nn with the pressure in kPa and the volume in litres, which are the units built into the given value of RR:

n=PVRT=(98.5 kPa)(5.00 L)(8.314 L kPa mol−1 K−1)(295.15 K)=492.52453.9=0.2007 moln = \frac{PV}{RT} = \frac{(98.5\ \text{kPa})(5.00\ \text{L})}{(8.314\ \mathrm{L\,kPa\,mol^{-1}\,K^{-1}})(295.15\ \text{K})} = \frac{492.5}{2453.9} = 0.2007\ \text{mol}

Convert to mass using the molar mass of O2\mathrm{O_2} (two oxygen atoms, not one; using 16.00 g/mol would give half the answer):

m=0.2007 mol×32.00 g/mol=6.42 gm = 0.2007\ \text{mol} \times 32.00\ \text{g/mol} = 6.42\ \text{g}

A quick sanity check: at 0 °C and 101.3 kPa one mole of ideal gas occupies 22.4 L, so 5.00 L at a slightly higher temperature and slightly lower pressure should hold a little under 0.22 mol, which 0.2007 mol is.

Mass of oxygen: 6.42 g.

Question 3Worked problemChemical equilibrium & acid-base chemistry

A 500.0 mL solution contains 0.200 M ethanoic (acetic) acid, CH3COOH\mathrm{CH_3COOH}, for which Ka=1.8×10−5K_a = 1.8\times10^{-5} at 25 °C.

(a) Calculate the pH of the solution and show that the usual small-xx approximation is justified.

(b) 1.60 g of solid sodium hydroxide (molar mass 40.00 g/mol) is dissolved in the solution. Assuming no change in volume, calculate the new pH.

(c) State, with a reason, whether the solution in (b) is a buffer.

Worked solution
Answer:

(a) pH = 2.72; (b) pH = 4.57; (c) yes, it contains comparable amounts of a weak acid and its conjugate base

(a) Set up the weak-acid equilibrium and let x=[H3O+]x = [\mathrm{H_3O^+}] at equilibrium:

CH3COOH+H2O⇌H3O++CH3COO−,Ka=x20.200−x\mathrm{CH_3COOH} + \mathrm{H_2O} \rightleftharpoons \mathrm{H_3O^+} + \mathrm{CH_3COO^-}, \qquad K_a = \frac{x^2}{0.200 - x}

Because KaK_a is small compared with the concentration, assume 0.200−x≈0.2000.200-x \approx 0.200:

x=(1.8×10−5)(0.200)=1.90×10−3 M,pH=−log⁡(1.90×10−3)=2.72x = \sqrt{(1.8\times10^{-5})(0.200)} = 1.90\times10^{-3}\ \text{M}, \qquad \text{pH} = -\log(1.90\times10^{-3}) = 2.72

The approximation is justified because xx is only 0.95%0.95\% of 0.200 M, well under the 5 % limit.

(b) NaOH is a strong base, so it reacts completely with the weak acid before any equilibrium is considered. Work in moles, because the neutralization is mole-for-mole:

nNaOH=1.6040.00=0.0400 mol,nCH3COOH=0.5000 L×0.200 M=0.1000 moln_{\mathrm{NaOH}} = \frac{1.60}{40.00} = 0.0400\ \text{mol}, \qquad n_{\mathrm{CH_3COOH}} = 0.5000\ \text{L} \times 0.200\ \text{M} = 0.1000\ \text{mol}

After the reaction CH3COOH+OH−→CH3COO−+H2O\mathrm{CH_3COOH} + \mathrm{OH^-} \rightarrow \mathrm{CH_3COO^-} + \mathrm{H_2O}, 0.0600 mol of acid remains and 0.0400 mol of ethanoate has formed. Both are in the same volume, so the mole ratio can be used directly in the Henderson–Hasselbalch equation, with pKa=−log⁡(1.8×10−5)=4.745\text{p}K_a = -\log(1.8\times10^{-5}) = 4.745:

pH=pKa+log⁡[A−][HA]=4.745+log⁡0.04000.0600=4.745−0.176=4.57\text{pH} = \text{p}K_a + \log\frac{[\mathrm{A^-}]}{[\mathrm{HA}]} = 4.745 + \log\frac{0.0400}{0.0600} = 4.745 - 0.176 = 4.57

(c) The solution now contains a weak acid and its conjugate base in comparable amounts (ratio 2 : 3), so it resists pH change when small amounts of acid or base are added: it is a buffer, and its pH lies within one unit of pKa\text{p}K_a, the useful buffering range.

(a) pH = 2.72; (b) pH = 4.57; (c) yes, the solution in (b) is a buffer.

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