Answer:
56.51L
Explanation:
Firstly, we need to write the equation of reaction. This means we are writing equation of reaction for the decomposition of sodium azide.
2NaN3 ——> 3N2. + 2Na
We can see from the equation that 2 moles of the side gives 3 moles of nitrogen.
We need to get the number of moles of the azide actually in the reaction.
This is gotten by dividing the mass of the azide by the molar mass. The molar mass of the azide is 23 + 3(14) = 23 + 42 = 65g/mol
The number of moles is 100g/65 = 1.54 moles
From here, we get the number of moles of nitrogen produced after decomposition.
From what we can see:
2 moles of azide = 3 nitrogen moles
1.54 moles of azide = x nitrogen moles
x = (1.54 * 3)/2 = 2.31 moles of nitrogen
To get the volume produced, we use the ideal gas equation
PV = nRT
P = 1 atm
V= ?
n = 2.31 moles
R = 0.08205 L atm mol^-1 K^-1
T = 25 + 273.15 = 298.15K
V = nRT/P= (2.31 * 0.08205 * 298.15)/1 = 56.51L