Balancing Equations
In a balanced chemical equation, this law states that the total mass of reactants equals the total mass of products.
What is the Law of Conservation of Mass?
This quantity — 6.022 × 10²³ — is the number of particles in one mole of any substance.
What is Avogadro’s number?
This gas law states that at constant temperature, the pressure and volume of a gas are inversely proportional: P₁V₁ = P₂V₂.
What is Boyle’s Law?
At STP, one mole of any ideal gas occupies this volume.
22.4 liters
This term describes a solution where water is the solvent.
What is an aqueous solution?
Balance this equation: C4H8 + O₂ →CO2 + H₂O. State the coefficients in order.
C4H8 + O₂ → CO2+ H₂O (coefficients: 6,4,4)
What is 6,4,4
How many moles are in 44 grams of CO₂? (Molar mass of CO₂ = 44 g/mol)
1 mole of CO₂
A gas occupies 3.0 L at 300 K. What volume does it occupy at 600 K at constant pressure?
6.0 L (Charles’s Law: V₁/T₁ = V₂/T₂ → 3.0/300 = V₂/600)
How many liters of O₂ at STP are needed to completely combust 1 mole of CH₄? (CH₄ + 2O₂ → CO₂ + 2H₂O)
44.8 L (2 mol O₂ × 22.4 L/mol)
Calculate the molarity of a solution made by dissolving 4.0 g of NaOH (molar mass = 40 g/mol) in enough water to make 500 mL of solution.
0.20 M (0.10 mol NaOH / 0.500 L)
In this type of reaction, a single compound breaks down into two or more simpler substances. Balance: CaCO₃ → CaO + CO₂.
Decomposition reaction. Already balanced: 1 CaCO₃ → 1 CaO + 1 CO₂
In the reaction N₂ + 3H₂ → 2NH₃, how many moles of NH₃ are produced from 4 moles of H₂?
2.67 moles of NH₃ (4 mol H₂ × 2 mol NH₃ / 3 mol H₂)
State the conditions for Standard Temperature and Pressure (STP).
0°C (273.15 K) and 1 atm (101.325 kPa)
How many grams of CO₂ are produced when 11.2 L of CO at STP reacts with excess O₂? (2CO + O₂ → 2CO₂; CO₂ = 44 g/mol)
22 g CO₂ (11.2 L ÷ 22.4 L/mol = 0.5 mol CO → 0.5 mol CO₂ × 44 = 22 g)
Unlike molarity, this concentration unit is defined as moles of solute per kilogram of solvent, and does not change with temperature.
What is molality (m)?
Balance this combustion reaction: C₃H₈ + O₂ → CO₂ + H₂O. What are the coefficients?
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O (coefficients: 1, 5, 3, 4)
This is the reactant that is completely consumed first and limits the amount of product formed in a reaction.
What is the limiting reactant (limiting reagent)?
Using the Ideal Gas Law (PV = nRT), find the pressure of 2.0 mol of gas at 300 K in a 10.0 L container. (R = 0.0821 L·atm/mol·K)
P = nRT/V = (2.0)(0.0821)(300)/10.0 = 4.93 atm
How many litres of H₂ gas at 27°C and 2.0 atm are produced from 0.5 mol Zn reacting with excess HCl? (Zn + 2HCl → ZnCl₂ + H₂; R = 0.0821)
6.16 L (PV = nRT → V = (0.5)(0.0821)(300)/2.0 = 6.16 L)
A solution contains 0.5 mol of glucose dissolved in 250 g of water. Calculate the molality.
2.0 m (0.5 mol / 0.250 kg)
Balance the following reaction:
C7H6O2 + O2 → CO2 + H2O.
2 C7H6O2+ 15O2 → 14CO2 + 6H₂O
What is 2, 15, 14, 6?
In a reaction, 5.0 g of H₂ reacts with excess O₂ to form water. Calculate the theoretical yield of H₂O. (H₂ = 2 g/mol, H₂O = 18 g/mol; reaction: 2H₂ + O₂ → 2H₂O)
45 g H₂O (5.0 g ÷ 2 g/mol = 2.5 mol H₂ → 2.5 mol H₂O × 18 g/mol = 45 g)
This law states that the total pressure of a gas mixture equals the sum of the partial pressures of each component gas.
What is Dalton’s Law of Partial Pressures?
N₂ + 3H₂ → 2NH₃. If 5.60 L of N₂ at STP reacts completely, what volume of NH₃ is produced at the same conditions?
11.2 L NH₃ (0.25 mol N₂ × 2 mol NH₃/1 mol N₂ = 0.5 mol × 22.4 L/mol)
A student prepares 250 mL of a 1.0 M NaCl stock solution. They need 100 mL of a 0.25 M solution. Using C₁V₁ = C₂V₂, what volume of stock solution is needed, and how much water is added?
25 mL of stock solution + 75 mL of water (1.0 × V₁ = 0.25 × 100 → V₁ = 25 mL)