GCSE-IGCSE-KS4 Calculations Part 7.1 Extra Mole Questions for GCSE-AS-A2-IB Chemistry s at Doc Brown's Chemistry Clinic docb4

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ANSWERS to Part 7 - the more advanced mole Q's

On-line chemistry CALCULATIONS for KS4 Science, GCSE Chemistry and AS Chemistry (basic calculations)

* ORIGINAL mole Q

If you find these useful or spot a silly error please EMAIL query?comment

Mole calculations introduction * Molar gas volume * Advanced Redox titration Q's * Non-redox titration Q's

Qa7.1 (a) f. mass Al₂O₃ = 102, 2 ÷ 102 x 3 x 6.02 x 10²³ = 3.54 x 10²² oxide ions

(b) f. mass H₂ = 2, 3 ÷ 2 x 6.02 x 10²³ = 9.03 x 10²³ molecules

(c) 1.2 ÷ 24000 x 6.02 x 10²³ = 3.01 x 10¹⁹ molecules

(d) f. mass Cl₂ = 71, 3 ÷ 71 x 6.02 x 10²³ = 2.54 x 10²² molecules

(e) Neon exists as single atoms (A_r = 20), 10 ÷ 20 x 6.02 x 10²³ = 3.01 x 10²³ atoms

(f) 2Na + 2H₂O ==> 2NaOH + H₂,

1 mole sodium gives 0.5 moles hydrogen,

mole Na = 0.2 ÷ 23 = 0.008696, so mole H₂ = 0.008696 ÷ 2 = 0.004348

so volume H₂ = 0.004348 x 24000 = 104.3 cm³ or 0.104 dm³

(g) e.g. Mg + 2HCl ==> MgCl₂ + H₂

1 mole magnesium gives 1 mole hydrogen, mole Mg = 2 ÷ 24.3 = 0.0823

so mole H₂ = 0.0823, so volume H₂ = 0.0823 x 24 = 1.975 dm³

(h) both 1 mole of Na₂CO₃ or NaHCO₃ will give 1 mole of CO₂

(1) V_CO2 = mol Na₂CO₃ x 24000 = 0.76 ÷ 106 x 24000 = 172cm³

(2) V_CO2 = mol NaHCO₃ x 24000 = 0.76÷ 84 x 24000 = 217cm³

(i) Zn + 2HCl ==> ZnCl₂ + H₂, mole H₂ = mole HCl ÷ 2

mol HCl = 50 ÷ 1000 x 0.2 = 0.01 mol

so mole H₂ = 0.005, V_H2 = 0.005 x 24000 = 120cm³

(j) CaCO₃ + 2HCl ==> CaCl₂ + H₂O + CO₂

mole CO₂ = mole HCl ÷ 2, mol HCl = 75 ÷ 1000 x 0.05 = 0.00375 mol

so mole CO₂ = 0.001875, VCO₂ = 0.001875 x 24000 = 45cm³