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?

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Stich

Chemistry

30 March, 13:35

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?

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Casey Rivas · Answer 1

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