Chemical Reactions, Stoichiometry, and Solutions
Chemical Reactions, Stoichiometry, and Solutions
This domain tests balancing equations, classifying reactions, mole calculations, limiting reactants, and solution concentration. Expect questions that combine a balanced equation with molarity or mass-to-mass conversion.
Core rules
- Balance equations by changing coefficients, never subscripts. Conservation of mass requires the same number of each atom on both sides.
- Reaction types: synthesis (A + B → AB), decomposition (AB → A + B), single replacement (A + BC → AC + B), double replacement (AB + CD → AD + CB), and combustion of a hydrocarbon with O2 to produce CO2 and H2O.
- One mole = 6.022 × 10^23 particles. Molar mass is the sum of atomic masses in g/mol. Use mole ratios from the balanced equation to convert between substances.
- Molarity (M) = moles of solute / liters of solution. Dilution: M1V1 = M2V2.
- For limiting reactant problems, convert each reactant to moles, divide by its coefficient, and the smallest value identifies the limiting reactant. Percent yield = (actual yield / theoretical yield) × 100%.
Worked example
How many milliliters of 0.500 M HCl are needed to neutralize 25.0 mL of 0.200 M NaOH? The balanced equation is HCl + NaOH → NaCl + H2O, a 1:1 mole ratio. Moles NaOH = 0.0250 L × 0.200 mol/L = 0.00500 mol. Therefore HCl needed = 0.00500 mol. Volume HCl = moles / M = 0.00500 mol ÷ 0.500 mol/L = 0.0100 L = 10.0 mL. The reasoning: convert volume to moles using molarity, use the balanced ratio, then convert moles back to volume.
Traps that cost marks
Do not use the volume of solvent when calculating molarity; use the total volume of solution. Do not forget to convert mL to L. Do not balance by changing subscripts, and do not ignore coefficients in mole ratios. In dilution, add solvent but keep moles of solute constant. In limiting reactant problems, the reactant with the smaller mass is not automatically limiting; compare moles adjusted by coefficients.