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More Mole Calculations Answers
- The reaction equation is:
Na 2 CO 3 (aq) + H 2 SO 4 (aq) → Na 2 SO 4 (aq) + CO 2 (g) + H 2 O (l)
So, 1 mole of H 2 SO 4 neutralises 1 mole of Na 2 CO 3 Moles of H 2 SO 4 = volume (dm^3 ) x concentration (mol dm–^3 ) = (41.2 x 10–^3 ) x 0.05 = 2.05 x 10–^3 mol So, moles of Na 2 CO 3 = 2.05 x 10–^3 mol Also, moles of Na 2 CO 3 = volume (dm^3 ) x concentration (mol dm–^3 ) = ( 2 5.0 x 10–^3 ) x concentration Therefore, 2.05 x 10–^3 = (25.0 x 10–^3 ) x concentration Or, concentration of solution = 2.05 x 10–^3 / 25.0 x 10–^3 = 0.082 mol dm–^3 Number of moles in the whole 250.0 cm^3 solution = volume (dm^3 ) x concentration (mol dm–^3 ) = ( 250 .0 x 10–^3 ) x 0.082 = 0.0205 mol Mass of Na 2 CO 3 = Mr (g mol–^1 ) x moles (mol) = 106 .0 x 0.0205 = 2.17 g (Mr of Na 2 CO 3 = (23.0 x 2) + 12.0 + (3 x 16.0) = 106 .0) % of Na 2 CO 3 = mass of Na 2 CO 3 (g) / mass of soda crystals (g) x 100 = 2.17 / 5.45 x 100 = 39.8% or 40% (rounding up) The washing soda crystals are 40% sodium carbonate
- The reaction equation is:
Na 2 CO 3 (aq) + 2 HCl (aq) → 2NaCl (aq) + CO 2 (g) + H 2 O (l)
So, 2 moles of HCl neutralises 1 mole of Na 2 CO 3 Moles of HCl = volume (dm^3 ) x concentration (mol dm–^3 ) = (24.65 x 10–^3 ) x 0. 100 = 2.465 x 10–^3 mol So, moles of Na 2 CO 3 = 2.465 x 10–^3 / 2 = 1.233 x 10–^3 mol Mass of Na 2 CO 3 = Mr (g mol–^1 ) x moles (mol) =106.0 x (1.233 x 10–^3 ) = 0.131 g (Mr of Na 2 CO 3 = (23.0 x 2) + 12.0 + (3 x 16.0) = 106.0) Mass of H 2 O in the sample = 0.352 – 0.131 = 0.221 g Moles of H 2 O = 0.221 / 18 = 12.278 x 10–^3 mol Mole ratio of Na 2 CO 3 : H 2 O = 1.233 x 10–^3 : 12.278 x 10–^3 = 1 : 9.9 58 = 10 (rounding up). So, x = 10, Na 2 CO 3 .10H 2 O
- Here are the reactions:
Na 2 CO 3 (aq) + HCl (aq) → NaHCO 3 (aq) + NaCl (aq)
NaHCO 3 (aq) + HCl (aq) → NaCl (aq) + CO 2 (g) + H 2 O (l)
Phenolphthalein changes from pink to colourless at the end of the first reaction, when the pH of the solution is 9, and so the reaction measured is:
CO 32 –^ (aq) + H+^ (aq) → HCO 3 –^ (aq)
Methyl orange changes from yellow to orange at the end of the second reaction, when the pH of the solution is 7, and so the reaction measured is:
HCO 3 –^ (aq) + H+^ (aq) → CO 2 (g) + H 2 O (l)
So, in the first stage, the CO 32 –^ (aq) in the mixture is converted into HCO 3 –^ (aq). Moles of HCl (aq) to convert CO 32 –^ (aq) to HCO 3 –^ (aq) = volume x concentration = (2 8. 3 x 10–^3 ) x 0.08 = 2.26 x 10–^3 mol Moles of CO 32 –^ (aq) = 2.26 x 10–^3 mol Concentration of CO 32 –^ (aq) in the solution = moles / volume = 2.26 x 10–^3 mol / 25.0 x 10–^3 = 9.04 x 10–^2 mol dm–^3
% of Fe in the wire = (1.89 / 2.45) x 100 = 77.1% = 77% (rounding up)
- Balancing the electrons in the two half-equations means that we need to multiple the Fe2+^ (aq) equation by 5. Combining the two half-equations gives:
MnO 4 –^ (aq) + 8 H+^ (aq) + 5 Fe2+^ (aq) → Mn2+^ (aq) + 4 H 2 O (l) + 5Fe3+^ (aq)
Moles of MnO 4 –^ (aq) = volume x concentration = (25. 0 x 10 –^3 ) x 0.0 200 = 5. 00 x 10 –^4 mol Moles of Fe2+^ (aq) = (5.00 x 10 –^4 ) x 5 = 2.50 x 10 –^3 mol Mass of Fe2+^ = (2.50 x 10 –^3 ) x 55.8 = 0.14 0 g % of Fe2+^ in the lawn sand = (0.140 / 3.00) x 100 = 4.67% = 5 % (rounding up)
- Moles of excess EDTA (aq) = volume x concentration = (16.4 x 10–^3 ) x (1.115 x 10–^2 ) = 1.83 x 10–^4 mol Moles of total EDTA (aq) = volume x concentration = (50.0 x 10–^3 ) x (1.00 x 10–^2 ) = 0.50 x 10–^3 mol Moles of EDTA that reacts with Al3+^ = (0.50 x 10–^3 ) – (1.83 x 10–^4 ) = 3.17 x 10–^4 mol Moles of Al3+^ in 25.0 cm^3 portion (1 mol of Al3+^ reacts with 1 mol of EDTA) = 3.17 x 10–^4 mol Moles of Al3+^ in 25 0 .0 cm^3 portion = (3.17 x 10–^4 ) x 10 = 3.17 x 10–^3 mol Mr of aluminium sulfate hydrate = mass (g) / moles (mol) = 2.00 / (3.17 x 10–^3 ) = 630.9 g mol–^1 Mr of Al 2 (SO 4 ) 3 = (2 x 27.0) + (3 x 32.1) + (12 x 16.0) = 342.3 g mol–^1 Mr of n H 2 O = 630.9 – 342.3 = 288.6 g mol–^1 n = 288.6 / 18.0 = 16 .0 = 16 (rounding up)
- Moles of total MnO 4 –^ (aq) = volume x concentration = (25.9 x 10–^3 ) x 0.02 = 0.5 18 x 10–^3 mol Using the half equations:
MnO 4 –^ (aq) + 8H+^ (aq) + 5e–^ → Mn2+^ (aq) + 4H 2 O (l)
Fe2+^ (aq) → Fe3+^ (aq) + e–
We can deduce that the number of moles of Fe3+^ = 5 x number of moles of MnO 4 –^ (aq) = 5 x (0.518 x 10–^3 ) = 2.59 x 10–^3 mol Hydroxylamine reacts with hydrochloric acid as follows:
H 2 NOH + HCl → H 3 N+OH Cl–
Concentration of the chloride salt in mol dm–^3 = Concentration (g dm–^3 ) / Mr (g mol–^1 ) = 3.60 / (4 x 1.01) + 14.0 + 16.0 + 35.5) = 3.60 / 69.5 = 0.0518 mol dm–^3 Moles of H 3 N+OH Cl–^ = volume x concentration = (25.0 x 10–^3 ) x 0.0518 = 1.30 x 10–^3 mol Ratio of Fe3+^ : H 3 N+OH Cl–^ = 2.59 x 10–^3 / 1.30 x 10–^3 = 1.99 = 2 (rounding up)
- The H+^ and HO–^ ions are attracted to cathode and anode, respectively
At the cathode: 2H+^ + 2e–^ → H 2
At the anode: 4HO–^ → O 2 + 2H 2 O + 4e–
The volume of hydrogen produced at the cathode is twice the volume of oxygen
produced at the anode (as 4H+^ + 4e–^ → 2H 2 )
Coulombs (C) = Amperes (A) x Seconds (s) = 0.0250 x 6.50 x 60 x 60 = 585 C (60 minutes in an hour, and 60 seconds in a minute) One mole of electrons = 96 500 C Moles of electrons = 585 / 96 500 = 6.06 x 10–^3 mol As two moles of electrons are required to produce H 2 , Moles of H 2 = 6.06 x 10–^3 / 2 = 3.06 x 10–^3 mol The molar volume of a gas at s.t.p. = 22.4 dm^3 Volume of H 2 = 22.4 x (3.06 x 10–^3 ) = 0.0679 dm^3 = 67.9 cm^3 Volume of O 2 = 67.9 / 2 = 34.0 cm^3
- The dissociation equation is:
BaF 2 (s) ⇌ Ba2+^ (aq) + 2F–^ (aq)
So, Ksp = [Ba2+][F–]^2 = 1.00 x 10–^6 Assuming [Ba2+] is x, as we have two F–^ ions for every Ba2+^ ion, we can rewrite the equation as: 1.00 x 10–^6 = (x)(2x)^2 = (x)(4x^2 )
Total moles of gas = 18 + 54 + 48 = 120 mol The fraction of N 2 is: x(N 2 ) = 18/ So, the partial pressure of N 2 = p(N 2 ) = 200 x (18/120) = 30 atm The fraction of H 2 is: x(H 2 ) = 54/ So, the partial pressure of H 2 = p(H 2 ) = 200 x ( 54 /120) = 9 0 atm The fraction of NH 3 is: x(NH 3 ) = 48/ So, the partial pressure of NH 3 = p(H 2 ) = 200 x ( 48 /120) = 8 0 atm Kp = p(NH 3 )^2 / p(N 2 ) x p(H 2 )^3 = 80^2 / (30 x 80^3 ) = 2.9 x 10–^4 atm–^2
- The reaction equation is:
HNO 2 (aq) + NaOH (aq) ⇌ NaNO 2 (aq) + H 2 O (l)
pKa of HNO 2 = – logKa = – log(4.5 x 10–^4 ) = 3. Moles of HNO 2 = ( 4 0.0 x 10–^3 ) x 1 = 40 x 10 –^3 mol Moles of NaOH = (20.0 x 10–^3 ) x 1 = 20 x 10 –^3 mol So, 50% of HNO 2 reacts to form the salt (NaNO 2 ) and 50% is left behind pH = pKa + log[salt]/[acid] = 3.35 + log [20 x 10–^3 ] / [20 x 10–^3 ] = 3.35 + log 1 = 3.35 + 0 = 3.
- The mass of the solution is 500 g Heat capacity of solution = mass of solution x specific heat capacity of solution = 0.500 kg x 4200 J kg–^1 K–^1 = 2100 J K–^1 Heat capacity of copper calorimeter = mass of calorimeter x specific heat capacity of copper = 0.500 kg x 400 J kg–^1 K–^1 = 200 J K–^1 Increase in temperature = 2.50 °C Heat evolved = (heat capacity of solution + heat capacity of copper) x T = (2100 + 200) x 2.50 = 2300 x 2.50 = 5750 J Moles of water formed = moles of NaOH = moles of HCl Moles of water formed = (250 x 10–^3 ) x 0.400 = 0.100 mol Heat evolved per mole of water = 5750 / 0.100 = 57 500 J mol–^1
Standard enthalpy of neutralisation = – 57.5 kJ mol–^1
- For compound A: Moles of carbon = 71.1 / 12.0 = 5.93 mol of C Moles of nitrogen = 10.4 / 14.0 = 0.743 mol of N Moles of oxygen = 11.8 / 16.0 = 0.738 mol of O Moles of hydrogen = 6.7 / 1.01 = 6.63 mol of H Dividing the numbers by the lowest number (0.74) gives C 8 H 9 NO Also, C 8 H 9 NO has an Mr = (12.0 x 8) + (1.01 x 9) + 14.0 +16 = 135. 1 g mol–^1 Which is consistent with the given Mr For compound B: Moles of carbon = 77.1 / 12.0 = 6.43 mol of C Moles of nitrogen = 15.1 / 14.0 = 1.08 mol of N Moles of hydrogen = 7.5 / 1.01 = 7.43 mol of H Dividing the numbers by the lowest number (1.08) gives C 6 H 7 N Also, C 6 H 7 N has an Mr = (12.0 x 6) + (1.01 x 7) + 14.0 = 93.07 g mol–^1 Which is consistent with the given Mr Bromine water is decolourised and forms a white precipitate on reaction with aniline (C 6 H 5 NH 2 ) or phenol (PhOH). So, C 6 H 7 N is likely to be aniline, which, being an aromatic amine is basic. Hydrolysis of A (C 8 H 9 NO) gives C 6 H 5 NH 2. So, A must contain the C 6 H 5 NH– fragment. This leaves a C 2 H 3 O group, likely to be – COCH 3 , as the amide C 6 H 5 NHCOCH 3 would be hydrolysed to C 6 H 5 NH 2 (B) and CH 3 CO 2 H. A = C 6 H 5 NHCOCH 3 B = C 6 H 5 NH 2
