The goal of this unit is to use the coefficients in a reaction equation as amount-of-substance ratios and freely calculate mass, volume, and number of molecules.
Basic Knowledge
The procedure for calculations using a chemical equation can be summarized in the following 3 steps.
① Convert the given quantity to mol (for mass, ÷ molar mass; for gas volume, ÷ 22.4)
② Calculate the mol of the desired substance from the coefficient ratio in the reaction equation
③ Convert mol back to the requested unit (g, L, number of particles, etc.)
Example: How many g of H2O are produced when 4.0 g of H2 reacts with sufficient O2?
① H2: 4.0 ÷ 2.0 = 2.0 mol
② From the equation 2H2 + O2 → 2H2O, H2O is also 2.0 mol
③ 2.0 × 18 = 36 g
Reaction with excess reactant (One reactant remains while the other is completely consumed. Calculate based on the smaller amount)
Limiting reactant (The substance consumed first that stops the reaction. Use it as the basis)
Complete combustion (Combustion of a hydrocarbon that produces CO2 and H2O)
3-step method (g/L/number of particles → mol → mol → g/L/number of particles)
Chemical reaction yield (The ratio of the actual amount produced to the theoretically produced amount. Expressed as %)
Further Exploration (Background and Significance)
The rule for calculations using chemical equations is to “go through mol.” Setting up proportional equations directly between g values or L values is easy to get wrong. Example of a reaction with excess reactant:
Problem: If 2.0 mol of H2 and 2.0 mol of O2 react, how many mol of H2O are produced?
From 2H2 + O2 → 2H2O, 1.0 mol of O2 is sufficient to completely react with 2.0 mol of H2. Therefore, 1.0 mol of O2 remains, and H2 is the limiting reactant. Thus, the amount of H2O is 2.0 mol.
These calculations are also important in industrial production and lead to reaction designs that minimize waste.
- Perform calculations through mol (3-step method)
- Coefficient ratio = amount-of-substance ratio
- When reactants are present in excess, use the limiting reactant as the basis
- g → mol: ÷ molar mass
- L (gas) → mol: ÷ 22.4 (at standard conditions)
- Number of particles → mol: ÷ 6.02×1023
- Check unit consistency after calculating
Points of Caution (Common Confusions)
① “Proportion directly between g values” is incorrect. Always go through mol. ② In a reaction with excess reactant, both reactants cannot necessarily be used up. The smaller amount determines the result. ③ When using gas volume, check whether the conditions are standard conditions. ④ In combustion problems, balance both H and C before determining the coefficients.
Practice
- When 12 g of carbon C undergoes complete combustion, how many g of CO2 are produced? (C=12, O=16)
- When 22.4 L of CH4 undergoes complete combustion at standard conditions, find the volume of O2 required.
- When 3.0 mol of H2 reacts with 1.0 mol of O2, how many mol of H2O are produced? Which gas remains?