The Mole & the Avogrado Constant - Practice Questions Flashcards
Apply the theory of The Mole & the Avogrado Constant
A sample contains 75 g of NaCl. The relative molecular mass (Mr) of NaCl is 58.5 g/mol. How many moles of NaCl are present?
Moles = 75 ÷ 58.5 = 1.28 mol
Calculate the mass of 2.5 moles of calcium carbonate (CaCO3), given that its Mr is 100 g/mol.
Mass = 2.5 × 100 = 250 g
A compound weighs 128 g and contains 4 moles. What is its relative molecular mass (Mr)?
Mr = 128 ÷ 4 = 32 g/mol
How many atoms are present in 1 mole of sodium (Na)?
6.02 × 10²³ atoms, as 1 mole = Avogadro’s number
A hydrated salt loses 5 g of water when heated. If the original mass was 20 g, how many grams of anhydrous salt remain?
20 - 5 = 15 g anhydrous salt
A sample of hydrated copper sulfate has a total mass of 15 g, and after heating, the anhydrous salt weighs 9 g. Calculate the mass of water lost.
15 - 9 = 6 g of water
The theoretical yield of a reaction is 3.2 g, but the actual yield is 2.5 g. Calculate the percentage yield.
(2.5 ÷ 3.2) × 100 = 78.1%
A reaction is expected to produce 200 g of product, but only 175 g is obtained. Find the percentage yield.
(175 ÷ 200) × 100 = 87.5%
A compound consists of 24 g of oxygen and has a total mass of 120 g. Calculate the percentage by mass of oxygen.
(24 ÷ 120) × 100 = 20%
Calculate the percentage by mass of iron in iron(III) oxide (Fe2O3). The relative atomic masses are Fe = 56 and O = 16.
Mass of Fe = 2 × 56 = 112
Total mass = (2 × 56) + (3 × 16) = 160
Percentage Fe = (112 ÷ 160) × 100 = 70%
A sample weighs 50 g, but only 45 g of it is pure substance. Calculate the percentage purity.
(45 ÷ 50) × 100 = 90%
A 30 g sample of a chemical contains 27 g of pure product. Determine its percentage purity.
(27 ÷ 30) × 100 = 90%
A sample contains 10 g of hydrogen and 80 g of oxygen. Find its empirical formula. The relative atomic masses are H = 1 and O = 16.
Moles of H = 10 ÷ 1 = 10
Moles of O = 80 ÷ 16 = 5
Ratio = 10:5, simplify to 2:1
Empirical formula = H2O
A compound has the empirical formula CH2 and a molecular relative formula mass (Mr) of 56. Find its molecular formula.
Empirical Mr = (12 × 1) + (1 × 2) = 14
Molecular Ratio = 56 ÷ 14 = 4
Molecular Formula = C4H8
Calculate the mass of magnesium oxide produced when 6 g of magnesium fully reacts with oxygen. The relative atomic masses are Mg = 24 and O = 16.
Moles of Mg = 6 ÷ 24 = 0.25 mol
Ratio Mg : MgO = 1:1, so 0.25 mol MgO is formed
Mass of MgO = 0.25 × (24 + 16) = 10 g
A reaction produces 5 moles of CO2. If the relative molecular mass of CO2 is 44 g/mol, find the mass of CO2 formed.
Mass = 5 × 44 = 220 g
9.2 g of sodium reacts with 8 g of sulfur to form sodium sulfide. Determine which reactant is limiting, given the relative atomic masses Na = 23 and S = 32.
Moles of Na = 9.2 ÷ 23 = 0.40 mol
Moles of S = 8 ÷ 32 = 0.25 mol
Ratio in equation = 2Na : 1S, so 0.40 moles Na requires 0.20 moles S
Since we have 0.25 moles of S, Na is limiting
A reaction requires 3 moles of substance A, but 4 moles are available. Is A the limiting reactant?
No, because more than the required amount is available, meaning A is in excess.
A solution contains 50 cm³ of 0.10 mol/dm³ sulfuric acid (H2SO4). How many moles of acid are present?
Convert cm³ to dm³ = 50 ÷ 1000 = 0.050 dm³
Moles = 0.10 × 0.050 = 0.005 mol
0.025 moles of sodium hydroxide (NaOH) is dissolved in 250 cm³ of water. Find the concentration in mol/dm³.
Convert cm³ to dm³ = 250 ÷ 1000 = 0.25 dm³
Concentration = 0.025 ÷ 0.25 = 0.10 mol/dm³
A student dissolves 10 g of NaOH in 2 dm³ of distilled water. Calculate the concentration in g/dm³.
Concentration = 10 ÷ 2 = 5 g/dm³
A solution of HCl has a concentration of 0.50 mol/dm³. What volume is required to neutralize 25 cm³ of 0.80 mol/dm³ KOH?
Convert cm³ to dm³ = 25 ÷ 1000 = 0.025 dm³
Moles of KOH = 0.80 × 0.025 = 0.02 mol
Ratio KOH : HCl = 1:1, so HCl also needs 0.02 mol
Volume = Moles ÷ Concentration = 0.02 ÷ 0.50 = 0.040 dm³ or 40 cm³
How many moles of gas occupy 48 dm³ at room temperature and pressure (RTP)?
Moles = Volume ÷ Molar Gas Volume
Moles = 48 ÷ 24 = 2 moles
A sample of 2 moles of oxygen gas (O2) is measured. How much volume does it occupy at RTP?
Volume = Moles × 24
Volume = 2 × 24 = 48 dm³
