Mol calculation questions
What mass of sulphur dioxide could be made by heating 1 tonne of ore which contained 50% by mass of pyrite (FeS₂)? 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
0.5 tonnes FeS2 produces 0.5/480 x 320 tonnes Fe2O3 and 0.5/480 x 512 tonnes SO2 Therefore (a) mass of Fe2O3 = 0.333 tonnes, and (b) mass of SO2 = 0.533 tonnes
Take the molar volume to be 24 dm³ at room temp. and pressure (rtp). Calculate the mass of 200cm³ of chlorine gas at rtp (RAM Cl = 35.5)
1 mol Cl2 weighs 71 g If 24000 cm3 (at rtp) weighs 71 g 200 cm3 weighs 200/24000 x 71 g = 0.592 g
Some dilute sulphuric acid, H₂SO₄, had a concentration of 4.9g dm⁻³. What is its concentration in mol dm⁻³ (RAMs: H = 1, O = 16, S = 32)
1 mol H2SO4 weighs 98 g 4.90 g is 4.9/98 mol = 0.0500 mol Concentration = 0.0500 mol dm-3 (0.0500 to show that the answer is accurate to 3 sig figs)
Take the molar volume to be 24 dm³ at room temp. and pressure (rtp). Calculate the volume occupied by 0.16g of oxygen at rtp (RAM O = 16)
1 mol O2 weighs 32 g So 32 g O2 occupies 24000 cm3 at rtp and 0.16 g O2 occupies 0.16/32 x 24000 cm3 at rtp = 120 cm3
What is the maximum mass of calcium carbonate which will react with 25 cm³ of if 2 mol dm⁻³ hydrochloric acid? (RAMs: C = 12, O = 16, Ca = 40) CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
25.0 cm3 of 2.00 mol dm-3 HCl contains 25.0/1000 x 2.00 mol = 0.0500 mol The equation shows that you only need half the number of moles of calcium carbonate as of hydrochloric acid. No of moles of CaCO3 = 1/2 x 0.0500 mol = 0.0250 mol I mole of CaCO3 weighs 100 g. 0.0250 mol CaCO3 weighs 0.0250 x 100 g = 2.50 g
What mass of iron(III) oxide could be made by heating 1 tonne of ore which contained 50% by mass of pyrite (FeS₂)? 4FeS₂ + 11O₂ → 2Fe₂O₃ + 8SO₂
4 mol FeS2 produces 2 mol Fe2O3 and 8 mol SO2 (4 x 120) g FeS2 produces (2 x 160) g Fe2O3 and (8 x 64) g SO2 480 g FeS2 produces 320 g Fe2O3 and 512 g SO2 Or, 480 tonnes FeS2 produces 320 tonnes Fe2O3 and 512 tonnes SO2 1 tonne of ore contains 0.5 tonnes FeS2
Titanium is manufactured by reacting titanium (IV) chloride and sodium. What mass of sodium is required to produce 1 tonne of titanium? (RAMs: Na = 23, Ti = 48) TiCl₄ + 4Na → Ti + 4NaCl
From the equation, 4 mol Na gives 1 mol Ti. Substituting masses: 4 x 23 g Na give 48 g Ti i.e. 92 g Na give 48 g Ti Because the ratio is bound to be the same 92 tonnes Na give 48 tonnes Ti. Therefore, 92/48 tonnes Na give 1 tonne Ti. Mass of Na needed = 1.92 tonnes
What is the concentration in g dm⁻³ of some potassium hydroxide, KOH, solution with a concentration of 0.2 mol dm⁻³? (RAMs: H = 1, O = 16, K = 39)
KOH is 0.200 mol dm-3 1 mol KOH weighs 56 g 0.200 mol weighs 0.200 x 56 g = 11.2 g Concentration = 11.2 g dm-3
Calculate the unknown concentration: 25 cm³ of 0.1 mol dm⁻³ sodium hydroxide was neutralised by 20 cm³ of dilute nitric acid of unknown concentration. NaOH + HNO₃ → 2NaNO₃ + H₂O
No of moles of NaOH = 25/1000 x 0.100 = 0.00250 mol The equation shows a 1:1 reaction. No of moles of HNO3 = 0.00250 mol That's in 20.0 cm3. Concentration of HNO3 = 1000/20.0 x 0.00250 mol dm-3 = 0.125 mol dm-3
What mass of barium sulphate would be produced by adding excess barium chloride solution to 20 cm³ if copper(II) sulphate solution of concentration 0.1 mol dm⁻³? (RAMs: O = 16, S = 32, Ba = 137) BaCl₂ + CuSO₄ → BaSO₄ + CuCl₂
No of moles of copper(II) sulphate = 20/1000 x 0.100 = 0.00200 mol Equation shows that 1 mol CuSO4 produces 1 mol BaSO4 No of moles BaSO4 formed = 0.00200 mol 1 mol BaSO4 weighs 233 g 0.00200 mol BaSO4 weighs 0.00200 x 233 g = 0.466 g
What mass of Na₂CO₃ would be dissolved in 100cm³ of solution in order to get a concentration of 0.1 mol dm⁻³? (RAMs: C = 12, O = 16, Na = 23)
Relative formula mass Na2CO3 = 106 So 1 mol Na2CO3 weighs 106 g 0.100 mol weighs 10.6 g To get a 0.100 mol dm-3 solution you would have to dissolve 10.6 g in 1 dm3 (1000 cm3) If you only wanted 100 cm3 of solution you would only need 1.06 g Na2CO3
Take the molar volume to be 24 dm³ at room temp. and pressure (rtp). If 1 dm³ of gas at rtp weighs 1.42g, calculate the mass of 1 mole of the gas
The mass of 1 mole is what would occupy 24 dm3 at rtp If 1 dm3 weighs 1.42 g, 24 dm3 weighs 24 x 1.42 g = 34.1g
