How many grams of NaN₃ must be decomposed to fill a 42.0 L airbag with nitrogen gas at a pressure of 1.05 atm and a temp of 29.5 degrees Celsius?
2NaN3 ==> 2Na + 3N2
Use PV - nRT and solve for n at the conditions listed.
Convert mols N2 needed to mols NaN3 needed. Then grams NaN3 = mols NaN3 x molar mass NaN3.
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To find the amount of grams of NaN₃ needed to fill the airbag, we need to use the ideal gas law. The ideal gas law equation is given by:
PV = nRT
Where:
P is the pressure of the gas in atm,
V is the volume of the gas in liters,
n is the number of moles of gas,
R is the ideal gas constant (0.0821 L·atm/(mol·K)),
T is the temperature of the gas in Kelvin.
We need to convert the given values for pressure, temperature, and volume to the appropriate units before using the ideal gas law equation.
1 atm = 101.325 kPa (kilopascals)
42.0 L = 42.0 L
1 L·atm/(mol·K) = 0.0821 L·atm/(mol·K)
29.5 °C = 29.5 + 273.15 K
Now we can calculate the number of moles of gas using the ideal gas law equation:
PV = nRT
n = (PV) / (RT)
Let's substitute the given values into the equation:
n = (1.05 atm * 42.0 L) / (0.0821 L·atm/(mol·K) * 302.65 K)
After calculating this expression, we will get the value of moles of gas (n).
Finally, to find the number of grams of NaN₃ decomposed, we can use the molar mass of NaN₃ (sodium azide), which is NaN₃ = 65.01 g/mol.
The number of grams can be calculated by:
grams = moles * molar mass
By substituting the calculated value of moles from the previous step into the equation grams = moles * molar mass, we can determine the number of grams of NaN₃ required to fill the airbag.