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Problème 1 Une quantité de 2.5 moles de dioxyde de soufre (SO₂) occupe un volume de 65 L dans certaines conditions expérimentales.

Chemistry

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Problème 1
Une quantité de 2.5 moles de dioxyde de soufre (SO2SO_2) occupe un volume de 65 L65 \text{ L} dans certaines conditions expérimentales.

a) Ce gaz est-il dans les conditions normales ? Justifiez.

  1. Calculate the molar mass of SO2SO_2.

    M(SO2)=M(S)+2×M(O)=32 g/mol+2×16 g/mol=64 g/molM(SO_2) = M(S) + 2 \times M(O) = 32 \text{ g/mol} + 2 \times 16 \text{ g/mol} = 64 \text{ g/mol}

  2. Calculate the volume 2.5 moles of SO2SO_2 would occupy at CNTP.

    VCNTP=n×VmCNTP=2.5 mol×22.4 L/mol=56 LV_{CNTP} = n \times V_m^{CNTP} = 2.5 \text{ mol} \times 22.4 \text{ L/mol} = 56 \text{ L}

  3. Compare the given volume with the CNTP volume.

    The given volume is 65 L65 \text{ L}, which is different from 56 L56 \text{ L}.
    The gas is not under normal conditions (CNTP) because its volume (65 L65 \text{ L}) is not equal to the volume it would occupy at CNTP (56 L56 \text{ L}).

    b) Quel devrait être son volume aux CNTP ?

  4. Calculate the volume for 2.5 moles at CNTP.

    VCNTP=n×VmCNTP=2.5 mol×22.4 L/molV_{CNTP} = n \times V_m^{CNTP} = 2.5 \text{ mol} \times 22.4 \text{ L/mol} VCNTP=56 LV_{CNTP} = \boxed{\text{}56 \text{ L}\text{}}

    c) Déterminez la masse de 40 L40 \text{ L} de ce gaz dans les conditions expérimentales.

  5. Calculate the molar volume under experimental conditions.

    Vm,exp=Vexpnexp=65 L2.5 mol=26 L/molV_{m,exp} = \frac{V_{exp}}{n_{exp}} = \frac{65 \text{ L}}{2.5 \text{ mol}} = 26 \text{ L/mol}

  6. Calculate the number of moles in 40 L40 \text{ L} under experimental conditions.

    n=40 LVm,exp=40 L26 L/mol=2013 moln = \frac{40 \text{ L}}{V_{m,exp}} = \frac{40 \text{ L}}{26 \text{ L/mol}} = \frac{20}{13} \text{ mol}

  7. Calculate the mass.

    Mass=n×M(SO2)=2013 mol×64 g/mol\text{Mass} = n \times M(SO_2) = \frac{20}{13} \text{ mol} \times 64 \text{ g/mol} Mass=128013 g≈98.46 g\text{Mass} = \frac{1280}{13} \text{ g} \approx \boxed{\text{}98.46 \text{ g}\text{}}

    d) Quel volume occuperaient 80 g80 \text{ g} de SO2SO_2 dans les conditions expérimentales ?

  8. Calculate the number of moles in 80 g80 \text{ g} of SO2SO_2.

    n=MassM(SO2)=80 g64 g/mol=1.25 moln = \frac{\text{Mass}}{M(SO_2)} = \frac{80 \text{ g}}{64 \text{ g/mol}} = 1.25 \text{ mol}

  9. Calculate the volume using the experimental molar volume.

    V=n×Vm,exp=1.25 mol×26 L/molV = n \times V_{m,exp} = 1.25 \text{ mol} \times 26 \text{ L/mol} V=32.5 LV = \boxed{\text{}32.5 \text{ L}\text{}}

    e) Quel volume occuperaient 64 g64 \text{ g} de SO2SO_2 aux CNTP ?

  10. Calculate the number of moles in 64 g64 \text{ g} of SO2SO_2.

    n=MassM(SO2)=64 g64 g/mol=1 moln = \frac{\text{Mass}}{M(SO_2)} = \frac{64 \text{ g}}{64 \text{ g/mol}} = 1 \text{ mol}

  11. Calculate the volume at CNTP.

    V=n×VmCNTP=1 mol×22.4 L/molV = n \times V_m^{CNTP} = 1 \text{ mol} \times 22.4 \text{ L/mol} V=22.4 LV = \boxed{\text{}22.4 \text{ L}\text{}}

    Problème 2
    Une quantité de 3 moles de dioxygène (O2O_2) occupe un volume de 72 L72 \text{ L} dans certaines conditions expérimentales.

    a) Ce gaz est-il dans les conditions normales ? Justifiez.

  12. Calculate the molar mass of O2O_2.

    M(O2)=2×M(O)=2×16 g/mol=32 g/molM(O_2) = 2 \times M(O) = 2 \times 16 \text{ g/mol} = 32 \text{ g/mol}

  13. Calculate the volume 3 moles of O2O_2 would occupy at CNTP.

    VCNTP=n×VmCNTP=3 mol×22.4 L/mol=67.2 LV_{CNTP} = n \times V_m^{CNTP} = 3 \text{ mol} \times 22.4 \text{ L/mol} = 67.2 \text{ L}

  14. Compare the given volume with the CNTP volume.

    The given volume is 72 L72 \text{ L}, which is different from 67.2 L67.2 \text{ L}.
    The gas is not under normal conditions (CNTP) because its volume (72 L72 \text{ L}) is not equal to the volume it would occupy at CNTP (67.2 L67.2 \text{ L}).

    b) Quel devrait être son volume aux CNTP ?

  15. Calculate the volume for 3 moles at CNTP.

    VCNTP=n×VmCNTP=3 mol×22.4 L/molV_{CNTP} = n \times V_m^{CNTP} = 3 \text{ mol} \times 22.4 \text{ L/mol} VCNTP=67.2 LV_{CNTP} = \boxed{\text{}67.2 \text{ L}\text{}}

    c) Quelle est la masse de 48 L48 \text{ L} de ce gaz dans les conditions expérimentales ?

  16. Calculate the molar volume under experimental conditions.

    Vm,exp=Vexpnexp=72 L3 mol=24 L/molV_{m,exp} = \frac{V_{exp}}{n_{exp}} = \frac{72 \text{ L}}{3 \text{ mol}} = 24 \text{ L/mol}

  17. Calculate the number of moles in 48 L48 \text{ L} under experimental conditions.

    n=48 LVm,exp=48 L24 L/mol=2 moln = \frac{48 \text{ L}}{V_{m,exp}} = \frac{48 \text{ L}}{24 \text{ L/mol}} = 2 \text{ mol}

  18. Calculate the mass.

    Mass=n×M(O2)=2 mol×32 g/mol\text{Mass} = n \times M(O_2) = 2 \text{ mol} \times 32 \text{ g/mol} Mass=64 g\text{Mass} = \boxed{\text{}64 \text{ g}\text{}}

    d) Quel volume occuperaient 80 g80 \text{ g} de O2O_2 dans les conditions expérimentales ?

  19. Calculate the number of moles in 80 g80 \text{ g} of O2O_2.

    n=MassM(O2)=80 g32 g/mol=2.5 moln = \frac{\text{Mass}}{M(O_2)} = \frac{80 \text{ g}}{32 \text{ g/mol}} = 2.5 \text{ mol}

  20. Calculate the volume using the experimental molar volume.

    V=n×Vm,exp=2.5 mol×24 L/molV = n \times V_{m,exp} = 2.5 \text{ mol} \times 24 \text{ L/mol} V=60 LV = \boxed{\text{}60 \text{ L}\text{}}

    e) Quel volume occuperaient 32 g32 \text{ g} de O2O_2 aux CNTP ?

  21. Calculate the number of moles in 32 g32 \text{ g} of O2O_2.

    n=MassM(O2)=32 g32 g/mol=1 moln = \frac{\text{Mass}}{M(O_2)} = \frac{32 \text{ g}}{32 \text{ g/mol}} = 1 \text{ mol}

  22. Calculate the volume at CNTP.

    V=n×VmCNTP=1 mol×22.4 L/molV = n \times V_m^{CNTP} = 1 \text{ mol} \times 22.4 \text{ L/mol} V=22.4 LV = \boxed{\text{}22.4 \text{ L}\text{}}

    Problème 3
    Éthane (C2H6C_2H_6). Dans des conditions expérimentales particulières de température et de pression, 0.25 mole d'éthane (C2H6C_2H_6) occupe un volume de 7.5 L7.5 \text{ L}.

    a) Calculez le volume molaire de l'éthane dans ces conditions.

  23. Calculate the molar volume under experimental conditions.

    Vm,exp=Vexpnexp=7.5 L0.25 molV_{m,exp} = \frac{V_{exp}}{n_{exp}} = \frac{7.5 \text{ L}}{0.25 \text{ mol}} Vm,exp=30 L/molV_{m,exp} = \boxed{\text{}30 \text{ L/mol}\text{}}

    b) Quel volume occuperaient 0.60 mole de ce gaz dans les mêmes conditions ?

  24. Calculate the volume using the experimental molar volume.

    V=n×Vm,exp=0.60 mol×30 L/molV = n \times V_{m,exp} = 0.60 \text{ mol} \times 30 \text{ L/mol} V=18 LV = \boxed{\text{}18 \text{ L}\text{}}

    c) Quelle est la masse de 15 L15 \text{ L} d'éthane pris à 35∘C35^\circ C sous la pression de 120 KPa120 \text{ KPa} ?

  25. Calculate the molar mass of C2H6C_2H_6.

    M(C2H6)=2×M(C)+6×M(H)=2×12 g/mol+6×1 g/mol=24+6=30 g/molM(C_2H_6) = 2 \times M(C) + 6 \times M(H) = 2 \times 12 \text{ g/mol} + 6 \times 1 \text{ g/mol} = 24 + 6 = 30 \text{ g/mol}

  26. Convert the temperature to Kelvin.

    T=35∘C+273.15=308.15 KT = 35^\circ C + 273.15 = 308.15 \text{ K}

  27. Use the ideal gas law (PV=nRTPV = nRT) to find the number of moles.

    Use R=8.314 L⋅kPa/(mol⋅K)R = 8.314 \text{ L} \cdot \text{kPa}/(\text{mol} \cdot \text{K}).

    n=PVRT=120 kPa×15 L8.314 L⋅kPa/(mol⋅K)×308.15 Kn = \frac{PV}{RT} = \frac{120 \text{ kPa} \times 15 \text{ L}}{8.314 \text{ L} \cdot \text{kPa}/(\text{mol} \cdot \text{K}) \times 308.15 \text{ K}} n=18002562.981 mol≈0.7023 moln = \frac{1800}{2562.981} \text{ mol} \approx 0.7023 \text{ mol}

  28. Calculate the mass.

    Mass=n×M(C2H6)=0.7023 mol×30 g/mol\text{Mass} = n \times M(C_2H_6) = 0.7023 \text{ mol} \times 30 \text{ g/mol} Mass≈21.07 g\text{Mass} \approx \boxed{\text{}21.07 \text{ g}\text{}}

    Problème 4
    Dans certaines conditions de température et de pression, 0.50 mole d'ammoniac (NH3NH_3) occupe un volume de 12 L12 \text{ L}.

    a) Calculez le volume molaire de l'ammoniac dans ces conditions.

  29. Calculate the molar volume under experimental conditions.

    Vm,exp=Vexpnexp=12 L0.50 molV_{m,exp} = \frac{V_{exp}}{n_{exp}} = \frac{12 \text{ L}}{0.50 \text{ mol}} Vm,exp=24 L/molV_{m,exp} = \boxed{\text{}24 \text{ L/mol}\text{}}

    b) Quel volume occuperaient 1.25 mole de ce gaz dans les mêmes conditions ?

  30. Calculate the volume using the experimental molar volume.

    V=n×Vm,exp=1.25 mol×24 L/molV = n \times V_{m,exp} = 1.25 \text{ mol} \times 24 \text{ L/mol} V=30 LV = \boxed{\text{}30 \text{ L}\text{}}

    c) Calculez la masse de 18 L18 \text{ L} d'ammoniac dans ces conditions.

  31. Calculate the molar mass of NH3NH_3.

    M(NH3)=M(N)+3×M(H)=14 g/mol+3×1 g/mol=17 g/molM(NH_3) = M(N) + 3 \times M(H) = 14 \text{ g/mol} + 3 \times 1 \text{ g/mol} = 17 \text{ g/mol}

  32. Calculate the number of moles in 18 L18 \text{ L} under experimental conditions.

    n=18 LVm,exp=18 L24 L/mol=0.75 moln = \frac{18 \text{ L}}{V_{m,exp}} = \frac{18 \text{ L}}{24 \text{ L/mol}} = 0.75 \text{ mol}

  33. Calculate the mass.

    Mass=n×M(NH3)=0.75 mol×17 g/mol\text{Mass} = n \times M(NH_3) = 0.75 \text{ mol} \times 17 \text{ g/mol} Mass=12.75 g\text{Mass} = \boxed{\text{}12.75 \text{ g}\text{}}

    d) Quel est le volume de 17 g17 \text{ g} de NH3NH_3 pris à 40∘C40^\circ C sous la pression de 3.5 atm3.5 \text{ atm} ?

  34. Calculate the number of moles in 17 g17 \text{ g} of NH3NH_3.

    n=MassM(NH3)=17 g17 g/mol=1 moln = \frac{\text{Mass}}{M(NH_3)} = \frac{17 \text{ g}}{17 \text{ g/mol}} = 1 \text{ mol}

  35. Convert the temperature to Kelvin.

    T=40∘C+273.15=313.15 KT = 40^\circ C + 273.15 = 313.15 \text{ K}

  36. Use the ideal gas law (PV=nRTPV = nRT) to find the volume.

    Use R=0.0821 L⋅atm/(mol⋅K)R = 0.0821 \text{ L} \cdot \text{atm}/(\text{mol} \cdot \text{K}).

    V=nRTP=1 mol×0.0821 L⋅atm/(mol⋅K)×313.15 K3.5 atmV = \frac{nRT}{P} = \frac{1 \text{ mol} \times 0.0821 \text{ L} \cdot \text{atm}/(\text{mol} \cdot \text{K}) \times 313.15 \text{ K}}{3.5 \text{ atm}} V=25.7088653.5 L≈7.345 LV = \frac{25.708865}{3.5} \text{ L} \approx \boxed{\text{}7.345 \text{ L}\text{}}
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