Question

In a laboratory, 655 mL of CO2 st 30.0 degrees Celsius and 16.1 psi is combined with 2.70 g of KO2. The two chemicals react in a reaction container at STP conditions.

4 KO2 (g) + 2 CO2 (g) ——> 3 O2 (g) + 2 K2CO3 (s)

a. How many moles of CO2 are present initially before the reaction?

b. How many moles of KO2 are present initially before the reaction?

c. What is the limiting reagent in the reaction?

d. How many moles of O2 are produced?

e. What is the resulting volume of O2?

f. How many moles of CO2 are left I reacted in the container?

g. How many moles of KO2 are left I reacted in the container?

Answer 1

a)

Use ideal gas equation for calculation of mole of gas

Ideal gas equation

PV = nRT             where, P = atm pressure= 16.1psi = 1.0955 atm,

V = volume in Liter = 655 ml = 0.655 L

n = number of mole = ?

R = 0.08205L atm mol^-1 K^-1 =Proportionality constant = gas constant,

T = Temperature in K = 0⁰C = 273.15+ 30 = 303.15 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (1.0955X 0.655)/(0.08205X 303.15) = 0.0288 mole

0.0288 mole of CO₂ at initialy

b)molar mass of KO₂ = 71.1 gm/mole then 2.70 gm of KO₂ = 2.70 / 71.1 = 0.03797 mole

mole of KO₂ = 0.03797 mole

c)according to reaction 4 mole KO₂ react with 2 mole CO₂ then to react with 0.0288 mole of CO₂ required KO₂ =

0.0288 X 4 / 2 = 0.0576 mole but KO₂ given only 0.03797 mole therefore

KO₂ is limiting reactant.

d)According to reaction 4 mole KO₂ produce 3 mole of O₂ then 0.03797 mole of KO₂ produce O₂ =

0.03797 X 3 / 4 = 0.0285 mole

0.0285 mole of O₂ produced

e)