What is the minimum volume of an alkali solution with a mass fraction of potassium hydroxide of 20%

What is the minimum volume of an alkali solution with a mass fraction of potassium hydroxide of 20% and a density of 1.19 g / ml is required to absorb all the carbon monoxide (IV) obtained by burning propane with a volume of 112 liters?

Given:

W (KOH) = 20% or 0.2

Ρ (KOH) = 1.19 g / ml

V (C3H8) = 112 l

+ O2

V solution (KOH) -?

1) We write down the reaction equation for the combustion of propane and calculate the volume of the formed carbon dioxide:

112 L x L

С3Н8 + 5 О2 = 3 СО2 + 4 Н2О

1 mol 3 mol

Vm = 22.4 l / mol Vm = 22.4 l / mol

V = 22.4 L V = 22.4 L * 3

X = 112 * 3 * 22.4 / 22.4

X = 336 l (CO2)

2) We write the equation for the reaction of the interaction of potassium hydroxide with carbon dioxide and calculate the mass of the alkali that has reacted:

X g 336 l

2 KOH + CO2 = K2CO3 + H2O

2 mol 1 mol

M = 56 g / mol Vm = 22.4 l / mol

m = 112 g V = 22.4 l

x = 336 * 112 / 22.4

x = 1680 g (KOH)

3) Calculate the mass of the alkali solution:

m (solution) = m (substance) / Ρ * W

m (KOH solution) = 1680 g / 0.2 * 1.19 g / ml

m (KOH solution) = 7059 ml or 7, 059 l

Answer: 7,059 liters of KOH solution.