Elf-contained self-rescue breathing devices convert CO2 into O2 according to the following reaction: 4KO2(s) + 2CO2(g) → 2K2CO3(s) + 3O2(g) How many grams of KO2 are needed to produce 100.0 L of O2 at 20.0 °C and 1.00 atm?

394.05g of KO2

Explanation:

First, let us calculate the number of moles of O2 produced. This is illustrated below:

V = 100L

P = 1atm

T = 20°C = 20 +273 = 293K

R = 0.082atm.L/K /mol

n =?

PV = nRT

n = PV /RT

n = (1 x 100)/(0.082 x 293)

n = 4.16moles of O2

4KO2 + 2CO2 → 2K2CO3 + 3O2

From the equation,

4moles of KO2 produced 3moles of O2.

Therefore, xmol of KO2 will produce 4.16moles of O2 i.e

xmol of KO2 = (4.16 x 4)/3 = 5.55moles.

Converting 5.55moles O KO2 to grams, we have:

Molar Mass of KO2 = 39 + (2x16) = 39 + 32 = 71g/mol

Mass of KO2 = 5.55 x 71 = 394.05g

Therefore, 394.05g of KO2 is needed for the reaction

sure im new too!

explanation:

the cold front replaces the warmer air at the ground level changing the wind to be breezy and denser.there is also an immediate drop in the temperature.