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Chemistry (Old) International Baccalaureate (IB) Practice Question: Automobile air bags inflate by a rapid decomposition reaction. One typical compound used is guanidinium nitrate, $C(NH2)...

Automobile air bags inflate by a rapid decomposition reaction. One typical compound used is guanidinium nitrate, C(NH2)3NO3C(NH2)3NO_3C(NH2)3NO3​, which decomposes very rapidly to form nitrogen, water vapour and carbon.

  1. Deduce the equation for the decomposition of guanidinium nitrate.
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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

Note

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

    Note

    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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Related topics

3.2 Periodic trends1.2 The mole concept1.3 Reacting masses and volumesS1.4 Counting particles by mass: The moleS1.5 Ideal gases

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