Question 6.9: Using the Ideal Gas Equation in Reaction Stoichiometry Calcu...

?  mol  N _2 = 75.0  g  NaN _3 \times \frac{1  mol  NaN _3}{65.01  g  NaN _3} \times \frac{3  mol  N _2}{2  mol  NaN _3} = 1.73  mol  N _2

P = 735  mmHg \times \frac{1  atm }{760  mmHg } = 0.967  atm

V = ?

n = 1.73 mol

R = 0.08206  atm  L  mol^{-1} K^{-1}

T = 26 °C + 273 = 299 K

V = \frac{nRT}{P} = \frac{1.73  mol  \times  0.08206  atm  L  mol^{-1} K^{-1} \times  299  K}{0.967  atm} = 43.9  L

Assess
75.0 g NaN_3 is slightly more than one mole (M ≈ 65 g/mol). From this amount of NaN_3 we should expect a little more than 1.5 mol N_2(g). At 0 °C and 1 atm, 1.5 mol N_2(g) would occupy a volume of 1.5 × 22.4 = 33.6 L. Because the temperature is higher than 0 °C and the pressure is lower than 1 atm, the sample should have a volume somewhat greater than 33.6 L.