Question

Small rocket engines, such as model rockets, contain KClO33 to provide oxygen for the combustion reaction fueling the rocket motor. The relevant reaction for KClO33 is: 2 KClO33(s) → 2 KCl(s) + 3 O22(g) If 117.0 grams of solid KClO33 are burned in one of these rocket motors, how many liters of gaseous O22 are produced at 1.18 atm of pressure and 550.0 ºC? Gaseous sulfur dioxide, a component of acid rain, is made as a product of the combustion of sulfur-containing fuels. Calculate the volume of sulfur dioxide produced at 22.4 °C and 0.925 atm by the combustion of 8902 g of fuel that contains 5.55% of sulfur by mass. S88 (s) + 8 O22(g)→ 8 SO22(g)

Answer 1

(1) 2 KClO₃(s) ? 2 KCl(s) + 3 O2(g)

Molar mass of KClO₃ = At.mass of K + At.mass of Cl + (3xAt.mass of O)

                                = 39+35.5+(3x16)

                                = 122.5 g/mol

So number of moles of KClO₃ , n = mass/molar mass

                                                = 117.0 g / ( 122.5(g/mol)

                                                = 0.955 moles

From the balanced reaction ,

2 moles of KClO₃ produces 3 moles of Oxygen

0.955 moles of KClO₃ produces M moles of oxygen

M = ( 0.955x3) / 2

    = 1.43 moles of Oxygen

Calculation of volume of Oxygen produced :-

We know that ideal gas equation is PV = nRT

Where

T = Temperature = 550.0 oC = 550.0+273 = 823 K

P = pressure = 1.18 atm

n = No . of moles = 1.43 mol

R = gas constant = 0.0821 L atm / mol - K

V= Volume of the gas = ?

Plug the values we get V = (nRT) / P

                                     = 81.9 L

Do the other in the similar manner