Question

What volume of O₂ (at 1.064 atm and 21.1 ^oC) is produced by the decomposition of 6.48 kg of HgO?

2 HgO(s) --> 2 Hg(l) + O₂(g)

V = ___ L

Automobile airbags are inflated by the rapid decomposition of sodium azide (NaN₃).

2 NaN₃(s) --> 2 Na(s) + 3 N₂(g)

What volume of N₂ gas, at 1.02 atm and 23.7 ^oC, is produced by the complete decomposition of 129.3 g of sodium azide?

V = ___ L

Calculate the pressure of a 0.00179 mol of CCl₄ vapor that occupies 23 L at 21.1 ^oC if the vapor is treated as a van der Waals gas. (a = 20.4 atm L² mol^-2, b = 0.1383 L mol^-1)

p = ___ Torr

Answer 1

1) Given, Balanced chemical reaction,

2HgO(s) 2Hg(l) + O₂(g)

Temperature(T) = 21.2^oC + 273.15 = 294.35 K

Pressure(P) = 1.064 atm

Mass of HgO = 6.48 Kg x ( 1000g /1kg) = 6480 g

Firstly, calculating the number of moles of HgO,

= 6480 g HgO x (1 mol /216.59 g )

= 29.92 mol of HgO

Now, using the calculated moles of HgO and the mole ratio from the balanced chemical reaction, calculating the number of moles of oxygen gas,

= 29.92 mol of HgO x ( 1 mole of O₂ / 2 moles of HgO)

= 14.96 mol of O₂

Now, we know, the ideal gas equation,

PV= nRT

Rearranging the equation,

V = nRT /P

V = (14.96 mol x 0.08206 L.atm/mol.K x 294.35 K) / 1.064 atm

V = 340. L

2)  Given, Balanced chemical reaction,

2NaN₃(s) 2Na(s) + 3N₂(g)

Temperature(T) = 23.7^oC + 273.15 = 296.85 K

Pressure(P) = 1.02 atm

Mass of NaN₃ = 129.3 g

Firstly, calculating the number of moles of NaN₃,

= 129.3 g NaN₃ x (1 mol / 65.001 g )

= 1.9892 mol of NaN₃

Now, using the calculated moles of NaN₃ and the mole ratio from the balanced chemical reaction, calculating the number of moles of nitrogen gas,

= 1.9892 mol of NaN₃ x ( 3 mole of N₂ / 2 moles of NaN₃)

= 2.9838 mol of N₂

Now, we know, the ideal gas equation,

PV= nRT

Rearranging the equation,

V = nRT /P

V = (2.9838 mol x 0.08206 L.atm/mol.K x 296.85 K) / 1.02 atm

V = 71.3 L

3) Given,

Moles of CCl₄ = 0.00179 mol

Volume(V) = 23 L

Temperature(T) = 21.1 ^oC + 273.15 = 294.25 K

a = 20.4 atm L² mol^-2

b = 0.1383 L mol^-1

We know, Van-der Waals equation,

(P-n²a/V²)(V-nb) = nRT

Rearranging the Van-der Waals equation,

P = (nRT/V-nb) - ( n²a /V²)

Substituting the known values,

P = [( 0.00179 mol x 0.08206 L.atm/mol.K x  294.25 K) / (23 L - 0.00179 mol x 0.1383 L mol^-1)] - [(0.00179 mol)² x 20.4 atm L² mol^-2 / (23 L)²]

P = 0.001879 - 1.23 x 10^-7

P = 0.00188 atm

Converting atm to Torr,

= 0.00188 x ( 760 Torr / 1 atm)

= 1.428 Torr

P = 1.4 Torr