How much sodium azide is needed (in g) to pack inside an airbag that needs to inflate to 47 L? You can assume the temperature is 25 °C. and pressure is 1 atm. Answer to zero decimal points.Ideal gas law: PV = nRTR = 0.08205 L·atm/mol·KMW(N2) = 28.02 g.mol-1MW(NaN3) = 65.02 g.mol-1
Question
How much sodium azide is needed (in g) to pack inside an airbag that needs to inflate to 47 L? You can assume the temperature is 25 °C. and pressure is 1 atm. Answer to zero decimal points.Ideal gas law: PV = nRTR = 0.08205 L·atm/mol·KMW(N2) = 28.02 g.mol-1MW(NaN3) = 65.02 g.mol-1
Solution
The reaction for the decomposition of sodium azide (NaN3) is:
2NaN3 -> 2Na + 3N2
From this equation, we can see that 2 moles of NaN3 produce 3 moles of N2.
We also know that the molar mass of sodium azide (NaN3) is 65.02 g/mol.
We can use the ideal gas law (PV = nRT) to find the number of moles of N2 needed to inflate the airbag to 47 L at 25 °C and 1 atm.
Step 1: Convert the temperature from °C to K. K = °C + 273.15 So, 25 °C = 25 + 273.15 = 298.15 K
Step 2: Substitute the values into the ideal gas law equation and solve for n (number of moles). 1 atm * 47 L = n * 0.08205 L·atm/mol·K * 298.15 K
Solving for n gives: n = (1 atm * 47 L) / (0.08205 L·atm/mol·K * 298.15 K) = 1.91 mol of N2
Step 3: Use the stoichiometry of the reaction to find the moles of NaN3 needed. From the balanced equation, 2 moles of NaN3 produce 3 moles of N2. So, 1.91 moles of N2 will require (2/3) * 1.91 = 1.27 moles of NaN3.
Step 4: Convert the moles of NaN3 to mass using its molar mass. Mass = Moles x Molar mass Mass = 1.27 mol x 65.02 g/mol = 82.6 g
So, approximately 83 grams (rounded to 0 decimal points) of sodium azide is needed to pack inside an airbag that needs to inflate to 47 L at 25 °C and 1 atm.
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