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Sodium azide is used in automobile airbags to provide a source of nitrogen gas for rapid inflation in an accident - HSC - SSCE Chemistry - Question 19 - 2010 - Paper 1

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Sodium azide is used in automobile airbags to provide a source of nitrogen gas for rapid inflation in an accident. The equation shows the production of nitrogen gas ... show full transcript

Worked Solution & Example Answer:Sodium azide is used in automobile airbags to provide a source of nitrogen gas for rapid inflation in an accident - HSC - SSCE Chemistry - Question 19 - 2010 - Paper 1

Step 1

Convert Volume of Nitrogen Gas to Moles

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Answer

To find the number of moles of nitrogen gas produced, we can use the Ideal Gas Law: PV=nRTPV = nRT Where:

  • P = pressure (100 kPa)
  • V = volume (40 L)
  • n = number of moles
  • R = ideal gas constant (8.314 J/(mol·K))
  • T = temperature (273 K for 0°C)

First, convert pressure from kPa to Pa: 100 kPa = 100,000 Pa.

Now, rearranging the ideal gas law: n=PVRTn = \frac{PV}{RT} Substituting in the values: n=(100,000Pa)(40L)(8.314J/(molK))(273K)n = \frac{(100,000 Pa)(40 L)}{(8.314 J/(mol·K))(273 K)}

Calculating: n1.77moln \approx 1.77 mol

Step 2

Determine Moles of Sodium Azide Required

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Answer

From the balanced equation: 2NaN33N22NaN_3 \rightarrow 3N_2 It shows that 2 moles of sodium azide produce 3 moles of nitrogen gas. Therefore, we can set up a ratio: 2 moles of NaN33 moles of N2=x moles of NaN31.77 moles of N2\frac{2 \text{ moles of } NaN_3}{3 \text{ moles of } N_2} = \frac{x \text{ moles of } NaN_3}{1.77 \text{ moles of } N_2}

Cross-multiplying gives: x=23×1.771.18 moles of NaN3x = \frac{2}{3} \times 1.77 \approx 1.18 \text{ moles of } NaN_3

Step 3

Convert Moles of Sodium Azide to Mass

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Answer

Now, we need to find the mass of sodium azide. The molar mass of sodium azide (NaN₃) can be calculated as:

  • Na: 22.99 g/mol
  • N: 14.01 g/mol
  • Total: 22.99 + (3 x 14.01) = 65.01 g/mol

To find the mass, we use the formula: mass=n×molar massmass = n \times molar \ mass Substituting in our values: mass=1.18extmoles×65.01 g/mol76.7gmass = 1.18 ext{ moles} \times 65.01 \text{ g/mol} \approx 76.7 g

Thus, the answer rounds to 76 g.

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