The combustion of propane occurs when propane interacts with oxygen gas to produce water and carbon dioxide in the following reaction:
c3h8 (g) + 5o2(g) => 3co2 (g) + 4h2o (l)
if 2 moles of propane c3h8 (g) react at 373 k in a volume of 2.5 l, what is the resulting pressure in the cylinder from the co (g) produced in the reaction?

             Pressure =  73.49 atm

Explanation:

The balance chemical equation is as follow,

                                   C₃H₈ + 5 O₂  →  3 CO₂ + 4 H₂O

According to balanced equation,

           1 mole of C₃H₈ on combustion gives  =  3 moles of CO₂

So,

    2 moles of C₃H₈ on combustion will give  =  X moles of CO₂

Solving for X,

                      X  =  2 moles of C₃H₈ × 3 moles of CO₂ ÷ 1 mole of C₃H₈

                      X =  6 moles of CO₂

Now, in second step we will calculate the the pressure exerted by CO₂ at 2.5 L volume and 373 K temperature. For this we will use Ideal gas equation assuming the gas is acting as an ideal gas. Therefore,

Data:

       Temperature  =  T = 373 K

       Volume  =  V  =  2.5 L

       Moles  =  n  =  6 mol   ∴ As calculated above.

       Gas Constant  =  R  =  0.0821 atm.L.mol⁻¹.K⁻¹

Formula Used:

                         P V = n R T

Solving for P,

                          P = n R T / V

Putting Values,

                          P = 6 mol × 0.0821 atm.L.mol⁻¹.K⁻¹ × 373 K ÷ 2.5 L

                         P =  73.49 atm

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3

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