Question

How many grams of sodium azide, NaN3 would be required to provide 26.6 L of N₂ gas at 22.0 degree celsius and 1.10 atm? the molar mass of NaN3 is 65.01 g/mol

2 NaN₃(s) = 2 Na(s) + 3 N₂(g)

Answer 1

To produce 26.6 L of N2 gas at 22.0 degrees Celsius and 1.10 atm, you would need 52.01 grams of sodium azide (NaN3).

Step-by-step explanation:

To determine the number of grams of sodium azide (NaN3) needed to produce 26.6 L of N2 gas at 22.0 degrees Celsius and 1.10 atm, we need to use stoichiometry.

First, we calculate the moles of N2 gas using the ideal gas law: PV = nRT. The volume is converted to liters and the temperature to Kelvin. Plugging in the values, we get:

n = (1.10 atm) * (26.6 L) / (0.0821 L.atm.mol-1.K-1) * (295 K)

n = 1.20 mol

Next, we use the stoichiometric ratio from the balanced equation: 2 NaN3 (s) --> 2 Na (s) + 3 N2 (g).

From the balanced equation, for every 3 moles of N2 gas produced, we need 2 moles of NaN3. So, the moles of NaN3 needed is calculated as:

moles of NaN3 = (1.20 mol N2) * (2 mol NaN3 / 3 mol N2) = 0.80 mol NaN3

Finally, we calculate the mass of sodium azide using its molar mass:

mass of NaN3 = (0.80 mol NaN3) * (65.01 g/mol NaN3) = 52.01 g NaN3