Answers to the more advanced level mole concept
calculations
Doc
Brown's Chemistry  GCSE/IGCSE/GCE (basic A level) Online Chemical Calculations
ANSWERS to Part 7
 the more advanced mole Q's
ORIGINAL
mole Q
Quantitative chemistry
calculations This page has the
answers to the more advanced mole based questions from
calculations section 7. for more
advanced chemistry students. Online practice exam chemistry CALCULATIONS and
solved problems for KS4 Science
GCSE/IGCSE CHEMISTRY and basic starter chemical calculations for A level AS/A2/IB.
These revision notes and practice questions on how to do chemical
calculations and worked examples should prove useful for the new AQA,
Edexcel and OCR GCSE (9–1) chemistry science courses.
Spotted any careless error?
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GCSE calculation not covered?
Mole
calculations introduction * Molar gas
volume * Advanced
Redox titration Q's
* Nonredox titration Q's
Qa7.1
(a) f. mass Al_{2}O_{3} = 102, 2
÷ 102 x 3 x 6.02 x 10^{23} = 3.54 x 10^{22} oxide ions
(b) f. mass H_{2} = 2, 3 ÷ 2 x 6.02 x 10^{23} =
9.03 x 10^{23} molecules
(c) 1.2 ÷ 24000 x 6.02 x 10^{23} =
3.01 x 10^{19} molecules
(d) f. mass Cl_{2} = 71, 3 ÷ 71 x 6.02 x 10^{23} =
2.54 x 10^{22} molecules
(e) Neon exists as single atoms (A_{r} = 20), 10
÷ 20 x 6.02 x 10^{23} = 3.01 x 10^{23} atoms
(f) 2Na + 2H_{2}O
==>
2NaOH + H_{2},
1 mole sodium gives 0.5 moles hydrogen,
mole Na = 0.2
÷ 23 = 0.008696, so mole H_{2} = 0.008696 ÷ 2 = 0.004348
so volume H_{2} = 0.004348 x 24000 =
104.3 cm^{3}
or 0.104 dm^{3}
(g) e.g. Mg + 2HCl ==> MgCl_{2} + H_{2}
1 mole magnesium gives 1 mole hydrogen, mole Mg = 2
÷ 24.3 = 0.0823
so mole H_{2} = 0.0823, so
volume H_{2} = 0.0823 x 24 =
1.975 dm^{3}
(h) both 1 mole of Na_{2}CO_{3} or NaHCO_{3} will give 1 mole of CO_{2}
(1) V_{CO2} = mol Na_{2}CO_{3} x 24000 = 0.76
÷ 106 x 24000 = 172cm^{3}
(2) V_{CO2} = mol NaHCO_{3} x 24000 = 0.76÷ 84
x 24000 = 217cm^{3}
(i) Zn + 2HCl
==> ZnCl_{2} + H_{2 }, mole H_{2} = mole HCl
÷ 2
mol HCl = 50 ÷ 1000 x 0.2 = 0.01 mol
so mole H_{2} = 0.005, V_{H2} = 0.005 x 24000 =
120cm^{3}
(j) CaCO_{3} + 2HCl
==>
CaCl_{2} + H_{2}O + CO_{2}
mole CO_{2} = mole HCl
÷ 2, mol HCl = 75 ÷ 1000 x 0.05 = 0.00375 mol
so mole CO_{2} = 0.001875, VCO_{2} = 0.001875 x 24000 =
45cm^{3}
Above is typical periodic table used in GCSE sciencechemistry specifications in
doing chemical calculations,
and I've 'usually' used these values in my exemplar calculations to cover most
syllabuses
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