NaN3 is used in some automobile air bags. The impact of a collision triggers the decomposition of NaN3 as follows: 2NaN3(s) to 2Na(s) + 3N2(g). The nitrogen gas produced quickly inflates the air bag between the driver and the windshield. Calculate the volume of N2 generated at 21 degree celsius and 823 mmHg by the decomposition of 60.0g of NaN3.
You have the balanced equation.
2. Convert 60.0 g NaN3 to moles. moles = grams/molar mass.
3. Using the coefficients in the balanced equation, convert moles NaN3 to moles N2.
4. Using moles N2 from #3, use PV = nRT (n = moles) to calculate volume at the specified conditions. Don't forget to use T in Kelvin.
36.9
Why did the nitrogen molecule go to school?
Because it wanted to be an airbag-istrate!
But in all seriousness, let's calculate the volume of N2 generated. To do this, we'll need to use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, let's convert the given temperature of 21 degrees Celsius to Kelvin. We add 273.15 to get 294.15 K.
Next, we need to calculate the number of moles of N2 produced. From the balanced equation, we can see that 2 moles of NaN3 decompose to produce 3 moles of N2. So, we need to find the number of moles of NaN3 first:
Molar mass of NaN3 = 22.99 g/mol + 14.01 g/mol + (3 * 16.00 g/mol) = 65.02 g/mol
Number of moles of NaN3 = mass / molar mass = 60.0 g / 65.02 g/mol = 0.922 mol
Since 2 moles of NaN3 produce 3 moles of N2, we can use a mole ratio to find the number of moles of N2:
Number of moles of N2 = (0.922 mol * 3 mol) / 2 mol = 1.383 mol
Finally, we can substitute the values into the ideal gas law to solve for volume:
PV = nRT
V = (nRT) / P
V = (1.383 mol * 0.0821 L·atm/(mol·K) * 294.15 K) / 823 mmHg
Remember that 1 atm = 760 mmHg, so we can convert mmHg to atm:
V = (1.383 mol * 0.0821 L·atm/(mol·K) * 294.15 K) / (823 mmHg * (1 atm / 760 mmHg))
Now let's calculate the volume:
To calculate the volume of N2 gas generated, we need to use the ideal gas law equation: PV = nRT, where:
- P is the pressure of the gas (823 mmHg in this case). We need to convert it to atm, so 823 mmHg ÷ 760 mmHg/atm = 1.082 atm.
- V is the volume of the gas that we want to find.
- n is the number of moles of the gas, which we can calculate using the molar mass of NaN3 (sodium azide) and the given mass of NaN3.
- R is the ideal gas constant (0.0821 L·atm/(mol·K)).
- T is the temperature in Kelvin. We need to convert celsius to Kelvin, so 21°C + 273.15 = 294.15 K.
First, let's calculate the number of moles of NaN3:
1. Find the molar mass of NaN3:
- Na has a molar mass of 22.99 g/mol.
- N has a molar mass of 14.01 g/mol.
- Multiply the molar mass of sodium (22.99 g/mol) by 1 because there are 2 sodium atoms in NaN3.
- Multiply the molar mass of nitrogen (14.01 g/mol) by 3 because there are 3 nitrogen atoms in NaN3.
- Add the values: (22.99 g/mol × 1) + (14.01 g/mol × 3) = 65.00 g/mol.
2. Calculate the number of moles:
- Divide the given mass of NaN3 (60.0 g) by the molar mass of NaN3 (65.00 g/mol):
- 60.0 g ÷ 65.00 g/mol = 0.9231 mol.
Now, we can calculate the volume of N2 gas using the ideal gas law equation:
PV = nRT
V = (nRT) / P
V = (0.9231 mol × 0.0821 L·atm/(mol·K) × 294.15 K) / 1.082 atm
Calculating this, we get:
V ≈ 21.58 L
Therefore, the volume of N2 gas generated at 21°C and 823 mmHg by the decomposition of 60.0g of NaN3 is approximately 21.58 liters.