C2Mg + O2 → 2MgO[1]
There are 2 Mg atoms and 2 O atoms on each side.
A has 2 Cl on the left and 1 on the right, B has 2 O on the left and 1 on the right, and D has 3 O on the left and 5 on the right.
A2.0 g of H2[1]
2.0 g of H2 is 1.0 mol of molecules, which is 2.0 mol of atoms.
4.0 g of He is 1.0 mol of atoms. 16.0 g of O2 and 14.0 g of N2 are 0.50 mol of molecules, which is 1.0 mol of atoms each.
B2.0 mol[1]
2.0 mol of N2 would need 6.0 mol of H2, but only 3.0 mol is present, so H2 is the limiting reagent.
3.0 mol of H2 gives 2/3 × 3.0 = 2.0 mol of NH3. Option A works from the N2, which is in excess.
C7.2 dm3[1]
Each C3H8 gives 3CO2, so 0.10 mol gives 0.30 mol of CO2.
0.30 × 24 = 7.2 dm3. Option D counts the water as a gas, but it is a liquid at room temperature.
Blow pressure, high temperature[1]
At low pressure the particles are far apart, so their own volume and the attractions between them hardly matter.
At high temperature the particles move fast enough to overcome the attractions.
4Al + 3O2 → 2Al2O3[1]
Ba2+(aq) + SO42−(aq) → BaSO4(s). Na+ and Cl− are spectator ions.[1]
2(14.0 + 4 × 1.0) + 32.1 + 4 × 16.0 = 132.1[1]
(69 × 63 + 31 × 65) ÷ 100 = 63.6[1]
M = 40.0, mass = 0.250 × 40.0 = 10.0 g[1]
0.10 × 2 = 0.20 mol of Cl−, which is 0.20 × 6.02 × 1023 = 1.2 × 1023 ions[1]
C: 85.7 ÷ 12.0 = 7.14, H: 14.3 ÷ 1.0 = 14.3, ratio 1 : 2, so the empirical formula is CH2 (mass 14).
56 ÷ 14 = 4, so the molecular formula is C4H8.[2]
nCaCO₃ = 25.0 ÷ 100.1 = 0.2498 mol = nCaO
mass = 0.2498 × 56.1 = 14.0 g[2]
nCO₂ = 1.76 ÷ 44.0 = 0.0400 mol = nCaCO₃, mass = 0.0400 × 100.1 = 4.004 g
purity = 4.004 ÷ 5.00 × 100% = 80.1%[2]
0.200 × 0.0250 = 0.00500 mol[1]
0.500 × 58.5 = 29.3 g dm−3[1]
n = 8.0 ÷ 32.0 = 0.25 mol, V = 0.25 × 24 = 6.0 dm3[1]
40 cm3: the volume ratio CH4 : O2 is 1 : 2.[1]
298 K[1]
2.50 × 10−4 m3[1]
2.03 × 105 Pa[1]
V = nRT ÷ p = 1.00 × 8.31 × 273 ÷ 1.00 × 105 = 0.0227 m3
= 22.7 dm3[2]
mole fraction of He = 0.5 ÷ 2.0 = 0.25
p(He) = 0.25 × 200 = 50 kPa[2]
Helium atoms have only very weak instantaneous dipole–induced dipole forces and a small volume.
Ammonia molecules attract one another by hydrogen bonds, so ammonia deviates more.[2]
AChlorine atoms can have different numbers of neutrons.[1]
Chlorine is a mixture of isotopes, chlorine-35 and chlorine-37, which differ in their numbers of neutrons.
Ar is the average mass of the atoms, weighted by abundance, so it is not a whole number. The other statements do not explain it.
C204.4[1]
Tl-205 makes up 100 − 30 = 70% of the atoms.
Ar = (30 × 203 + 70 × 205) ÷ 100 = 204.4
D80%[1]
Let carbon-12 be x%: 12x + 13(100 − x) = 100 × 12.2, so 1300 − x = 1220.
x = 80%. The average lies nearer 12, so carbon-12 is the more common isotope.
A84 g/mol[1]
V has 13 protons and nucleon number 27: aluminium, which forms Al3+. W has 9 protons and nucleon number 19: fluorine, which forms F−.
The compound is VW3 (AlF3): 27 + 3 × 19 = 84 g/mol.
Cmethane, because it has a lower molecular mass[1]
Methane, CH4 (Mr = 16), is lighter than propene, C3H6 (Mr = 42), so its molecules move faster at the same temperature and it diffuses faster.
The rate of diffusion depends on the mass of the particles, not on how reactive the gas is or on its boiling point.
B0.3 mol of ammonia gas[1]
Each NH3 has 4 atoms: 0.3 × 4 = 1.2 mol of atoms, and 1.2 × 6.02 × 1023 = 7.2 × 1023.
A is 0.2 mol of atoms, C is 0.8 mol and D is 3.6 mol.
B2Lx[1]
1.0 g of Q is 1/x mol of molecules, which is L/x molecules.
Each molecule has 2 atoms, so there are 2L/x atoms.
BRuO2[1]
Mass of Ru = 5.79 − 1.39 = 4.40 g. nRu = 4.40 ÷ 101.1 = 0.0435 mol and nO = 1.39 ÷ 16.0 = 0.0869 mol.
Ru : O = 1 : 2, so the empirical formula is RuO2.
BCaFe2O4[1]
Oxygen is 100 − 18.5 − 51.9 = 29.6%. In 100 g: Ca 18.5 ÷ 40.1 = 0.461 mol, Fe 51.9 ÷ 55.8 = 0.930 mol, O 29.6 ÷ 16.0 = 1.85 mol.
Dividing by 0.461 gives 1 : 2 : 4, so CaFe2O4.
Brow B[1]
In 100 g: C 40.0 ÷ 12.0 = 3.33 mol, H 6.70 ÷ 1.0 = 6.70 mol, O 53.3 ÷ 16.0 = 3.33 mol. The ratio 1 : 2 : 1 gives CH2O, mass 30.
60 ÷ 30 = 2, so the molecular formula is C2H4O2.
ACH4[1]
nCO₂ = 11.0 ÷ 44.0 = 0.250 mol, so 0.250 mol of C. nH₂O = 9.0 ÷ 18.0 = 0.50 mol, so 1.0 mol of H.
C : H = 0.250 : 1.0 = 1 : 4, so the hydrocarbon is CH4.
B74%[1]
2Mg + O2 → 2MgO, so nMgO = nMg = 4.8 ÷ 24.3 = 0.198 mol, a theoretical yield of 0.198 × 40.3 = 7.96 g.
Percentage yield = 5.9 ÷ 7.96 × 100% = 74%.
D42.5 g[1]
60.0 g of urea is 1.00 mol. At 80% yield, the ammonia must be enough to make 1.00 ÷ 0.80 = 1.25 mol in theory.
2 mol of NH3 gives 1 mol of urea, so nNH₃ = 2.50 mol and its mass is 2.50 × 17 = 42.5 g.
B50%[1]
nNaHCO₃ = 42.0 ÷ 84 = 0.500 mol, so in theory 0.250 mol of CO2 forms, which is 0.250 × 24 = 6.00 dm3.
Percentage yield = 3.00 ÷ 6.00 × 100% = 50%.
C1.12 g/dm3[1]
Iron(III) sulfate is Fe2(SO4)3, so each unit gives 2 Fe3+: [Fe3+] = 0.02 mol/dm3.
In g/dm3: 0.02 × 55.8 = 1.12 g/dm3.
CC4H8[1]
n = 12 ÷ 24 = 0.50 mol, so Mr = 28 ÷ 0.50 = 56.
C4H8: 4 × 12 + 8 × 1 = 56.
B1.5 mol of C4H8[1]
144 dm3 of CO2 is 144 ÷ 24 = 6.0 mol. Each C atom in the fuel ends up in one CO2.
1.5 mol of C4H8 has 1.5 × 4 = 6.0 mol of C. A gives 3.0 mol, C gives 9.0 mol and D gives 12 mol.