sodium azide, decomposes according to the following equation: 2NaN3 (s) -> 2Na (s) + 3N2(g)

What mass of sodium azide is required to provide nitrogen needen to inflate a 80.0 L bag to a pressure to 1.35 atm at 25 c?

Use PV = nRT and calculate n, number of moles.

Using the coefficients in the balanced equation, convert moles N2 to moles NaN3.

Now convert moles NaN3 to grams. g = moles x molar mass.

To find the mass of sodium azide required, we need to use the ideal gas law and the balanced chemical equation for the decomposition of sodium azide.

1. First, let's convert the given volume from liters to moles. We can use the ideal gas law, which states: 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.

Given:
Pressure (P) = 1.35 atm
Volume (V) = 80.0 L
Temperature (T) = 25 °C = 298 K (since 0 °C is 273 K)

Using the ideal gas law, we can calculate the number of moles:
n = PV / RT
= (1.35 atm) * (80.0 L) / (0.0821 atm·L/mol·K * 298 K)
= 4.82 moles

2. Now, we need to convert moles of nitrogen to moles of sodium azide using the balanced chemical equation.

From the balanced equation: 2NaN3 (s) -> 2Na (s) + 3N2(g)
We can see that it takes 2 moles of sodium azide to produce 3 moles of nitrogen gas.

Since we want to inflate the bag with nitrogen, the number of moles of sodium azide required will be:
moles of sodium azide = (4.82 moles of N2) * (2 moles of NaN3 / 3 moles of N2)
= 3.21 moles of NaN3

3. Finally, we can calculate the mass of sodium azide required using its molar mass. The molar mass of sodium azide (NaN3) is the sum of the atomic masses of its individual elements:
Molar mass of NaN3 = (22.99 g/mol) + (14.01 g/mol * 3)
= 65.01 g/mol

Mass of sodium azide = (3.21 moles of NaN3) * (65.01 g/mol)
= 208.86 g

Therefore, approximately 208.86 grams of sodium azide are required to provide the nitrogen needed to inflate the 80.0 L bag to a pressure of 1.35 atm at 25 °C.