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    Automobile air bags inflate by a rapid decomposition reaction. One typical compound used is guanidinium nitrate, $C(NH_2)_3NO_3$, which decomposes very rapidly to form nitrogen, water vapour and carbon.

    Question
    SLPaper 2

    Automobile air bags inflate by a rapid decomposition reaction. One typical compound used is guanidinium nitrate, C(NH2)3NO3C(NH_2)_3NO_3C(NH2​)3​NO3​, which decomposes very rapidly to form nitrogen, water vapour and carbon.

    1.

    Deduce the equation for the decomposition of guanidinium nitrate.

    [1]
    Verified
    Solution

    C(NH2)3NO3 (s)→2N2 (g)+3H2O (g)+C (s)C(NH_2)_3NO_3 (s) → 2N_2 (g) + 3H_2O (g) + C (s)C(NH2​)3​NO3​ (s)→2N2​ (g)+3H2​O (g)+C (s) 1 mark

    2.

    Calculate the total number of moles of gas produced from the decomposition of 10.0 g of guanidinium nitrate.

    [2]
    Verified
    Solution
    1. Calculate the moles of guanidinium nitrate:
    n(C(NH2)3NO3)=10.0 gMr(C(NH2)3NO3)=10.0 g122.09 g mol−1=0.0819 moln({C(NH_2)_3NO_3}) = \frac{10.0\,\mathrm{g}}{\mathrm{M_r}({C(NH_2)_3NO_3})} = \frac{10.0\,\mathrm{g}}{122.09\,\mathrm{g\,mol^{-1}}} = 0.0819\,\mathrm{mol}n(C(NH2​)3​NO3​)=Mr​(C(NH2​)3​NO3​)10.0g​=122.09gmol−110.0g​=0.0819mol
    1. From the balanced equation, 1 mol of C(NH2)3NO3{C(NH_2)_3NO_3}C(NH2​)3​NO3​ produces 2 mol of N2{N_2}N2​ and 3 mol of H2O{H_2O}H2​O:
    n(N2)=0.0819 mol×2=0.164 moln({N_2}) = 0.0819\,\mathrm{mol} \times 2 = 0.164\,\mathrm{mol}n(N2​)=0.0819mol×2=0.164mol n(H2O)=0.0819 mol×3=0.246 moln({H_2O}) = 0.0819\,\mathrm{mol} \times 3 = 0.246\,\mathrm{mol}n(H2​O)=0.0819mol×3=0.246mol
    1. Total moles of gas produced = n(N2)+n(H2O)n({N_2}) + n({H_2O})n(N2​)+n(H2​O) = 0.164 mol + 0.246 mol = 0.410 mol

    Award 1 mark for correct substitution of values and 1 mark for the correct final answer.

    3.

    Calculate the pressure, in kPa, of this gas in a 10.0 dm3dm^3dm3 air bag at 127°C, assuming no gas escapes.

    [2]
    Verified
    Solution
    1. Convert the volume to SI units:

      V=10.0 dm3=10.0×10−3 m3=0.010 m3V = 10.0\,\mathrm{dm^3} = 10.0 \times 10^{-3}\,\mathrm{m^3} = 0.010\,\mathrm{m^3}V=10.0dm3=10.0×10−3m3=0.010m3
    2. Convert the temperature to Kelvin:

      T=127∘C+273.15=400.15 KT = 127^\circ\mathrm{C} + 273.15 = 400.15\,\mathrm{K}T=127∘C+273.15=400.15K
    3. Use the ideal gas equation to calculate the pressure:

      pV=nRTpV = nRTpV=nRT

      Rearranging for ppp:

      p=nRTVp = \frac{nRT}{V}p=VnRT​

      Substituting the values:

      p=(0.4095 mol)(8.314 J mol−1 K−1)(400.15 K)0.010 m3p = \frac{(0.4095\,\mathrm{mol})(8.314\,\mathrm{J\,mol^{-1}\,K^{-1}})(400.15\,\mathrm{K})}{0.010\,\mathrm{m^3}}p=0.010m3(0.4095mol)(8.314Jmol−1K−1)(400.15K)​ p=1.36×105 Pa=136 kPap = 1.36 \times 10^5\,\mathrm{Pa} = 136\,\mathrm{kPa}p=1.36×105Pa=136kPa

      Award 1 mark for correct substitution of values and 1 mark for the correct final answer

    4.

    Suggest why water vapour deviates significantly from ideal behaviour when the gases are cooled, while nitrogen does not.

    [2]
    Verified
    Solution

    Any two of:
    nitrogen non-polar/London/dispersion forces AND water polar/H-bonding 1 mark

    water has «much» stronger intermolecular forces 1 mark

    water molecules attract/condense/occupy smaller volume «and therefore deviate from ideal behaviour» 1 mark

    2 marks max

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    5.

    Another airbag reactant produces nitrogen gas and sodium.

    Suggest, including an equation, why the products of this reactant present a safety hazard.

    [2]
    Verified
    Solution

    2H2+O2→2H2O2 H_2 + O_2 \to 2 H_2O 2H2​+O2​→2H2​O

    OR

    4Na+2O2→2Na2O4Na + 2 O_2 \to 2 Na_2O4Na+2O2​→2Na2​O 1 mark

    hydrogen explosive
    OR
    highly exothermic reaction
    OR
    sodium reacts violently with water
    OR
    forms strong alkali 1 mark

    2 marks in total

    NOTE:Accept the equation of combustion of hydrogen.
    Do not accept just “sodium is reactive/dangerous”.

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