Stoichiometry (College Board AP® Chemistry): Flashcards
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- FrontConservation of Mass
Define the Law of Conservation of Mass.
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Cards in this collection (25)
Define the Law of Conservation of Mass.
Mass cannot be created or destroyed — the total mass of the reactants in a chemical reaction equals the total mass of the products.
True or False?
During a chemical reaction, atoms are created from energy to form new products.
False.
Atoms cannot be created or destroyed in a chemical reaction — they are rearranged to form new substances, so the total mass remains constant.
In the reaction 2H2 + O2 → 2H2O, if 4 g of H2 reacts and 36 g of H2O is produced, then .......... g of O2 was consumed.
In the reaction 2H2 + O2 → 2H2O, if 4 g of H2 reacts and 36 g of H2O is produced, then 32 g of O2 was consumed.
(Mass of reactants = mass of products: 4 + x = 36, so x = 32 g.)
How can the Law of Conservation of Mass be used to find an unknown reactant or product mass?
Set the total mass of reactants equal to the total mass of products, substitute the unknown as x, then solve the resulting equation. The sum of all known masses on one side determines the missing mass on the other.
True or False?
A balanced chemical equation is consistent with the Law of Conservation of Mass.
True.
A balanced equation has the same number of atoms of each element on both sides, which means the total mass of reactants equals the total mass of products — directly expressing conservation of mass.
Define molar mass.
The mass of one mole of a substance, expressed in g mol-1. It is numerically equal to the relative atomic mass (for elements) or relative molecular mass (for compounds), and is calculated using the periodic table.
The number of moles (n) in a sample is calculated using n = .......... ÷ M, where M is the molar mass.
The number of moles (n) in a sample is calculated using n = m ÷ M, where M is the molar mass.
(m = mass of the sample in grams.)
In the equation 2H2 + O2 → 2H2O, how many moles of H2O are produced when 6.7 mol of O2 reacts with excess H2?
13.4 mol of H2O.
The mole ratio of O2 to H2O is 1:2, so 6.7 mol O2 × (2 mol H2O / 1 mol O2) = 13.4 mol H2O.
Define limiting reactant.
The reactant that is completely consumed first in a chemical reaction, thereby determining the maximum amount of product that can be formed (the theoretical yield).
True or False?
The theoretical yield is the amount of product actually collected in the laboratory.
False.
The theoretical yield is the maximum amount of product predicted from the limiting reactant calculation. The amount actually collected is the actual yield, which is always less than or equal to the theoretical yield.
How is percent yield calculated, and what does it indicate?
Percent yield = (actual yield ÷ theoretical yield) × 100%.
It indicates how efficiently a reaction converts reactants into products under real laboratory conditions — always ≤ 100%.
In a limiting reactant calculation, the reactant that produces the .......... amount of product is identified as the limiting reactant.
In a limiting reactant calculation, the reactant that produces the least amount of product is identified as the limiting reactant.
Define molarity (M).
Molarity is the concentration of a solution expressed as moles of solute per litre of solution: M = n (mol) ÷ V (L). Units are mol L-1, written as M.
The ideal gas equation is PV = .......... RT, where n is the amount of gas in moles.
The ideal gas equation is PV = n RT, where n is the amount of gas in moles.
What unit of temperature must be used in the ideal gas equation, and how is it converted from °C?
Temperature must be in Kelvin (K). Convert using: K = °C + 273.
True or False?
In a solution calculation, you can calculate moles of solute by multiplying molarity by volume in mL.
False.
Molarity is defined as moles per litre, so volume must be converted to litres before multiplying: n = M × V (in L). Using mL directly gives an answer 1000 times too large.
In a stoichiometry problem involving a gas, what general sequence of steps links a given mass of reactant to a volume of gaseous product?
Convert mass of reactant to moles using molar mass.
Use the mole ratio from the balanced equation to find moles of gaseous product.
Apply PV = nRT (with T in K) to calculate the volume of gas.
When pressure is in atm and volume in L, the value of the gas constant R is .......... L atm mol-1 K-1.
When pressure is in atm and volume in L, the value of the gas constant R is 0.08206 L atm mol-1 K-1.
Define analyte.
The substance of unknown concentration in a titration — it is placed in the conical flask and its concentration is determined using the titrant.
Define equivalence point.
The point in a titration at which the moles of titrant added have reacted completely with the moles of analyte — neither reactant is in excess.
In a titration, the .......... is a solution of known concentration that is added from the burette to react with the analyte.
In a titration, the titrant is a solution of known concentration that is added from the burette to react with the analyte.
True or False?
In a standard titration setup, the analyte is placed in the burette and the titrant is in the conical flask.
False.
The burette is filled with the titrant (known concentration). The analyte (unknown concentration) is measured into the conical flask beneath the burette.
In an acid-base titration, a 30.0 mL HCl sample of unknown concentration required 25.0 mL of 0.150 M NaOH to reach the equivalence point. What is the concentration of the HCl?
0.125 M.
mol NaOH = 0.150 M × 0.0250 L = 0.00375 mol.
Mole ratio HCl:NaOH = 1:1, so mol HCl = 0.00375 mol.
[HCl] = 0.00375 mol ÷ 0.0300 L = 0.125 M.
What is the role of an indicator in a titration, and when should the burette tap be closed?
An indicator changes colour near the equivalence point to signal the endpoint. The burette tap is closed when one drop causes a permanent colour change in the analyte solution.
True or False?
Near the equivalence point, titrant should be added dropwise to avoid overshooting.
True.
As the equivalence point approaches, each drop has a large effect on the composition of the analyte solution. Adding dropwise gives precise control and prevents adding excess titrant past the endpoint.
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