Determine the mass of water obtained by the interaction of sodium hydroxide and sulfur oxide with a volume of 44.8 liters.

Let’s start by recording given:

V (SO₃) = 44.8 liters.

NaOH;

Vm = 22.4 L / mol.

Find:

m (H₂O) -?

Let’s start by determining the amount of SO3 in a volume of 22.4 liters:

n (SO₃) = V (SO₃): Vm = 44.8 k: 22.4 L / mol = 2 mol.

Let’s compose the reaction equation:

2NaOH + SO₃ = Na₂SO₄ + H₂O.

As a result of the interaction of 1 mol of SO3 with NaOH, 1 mol of H₂O is formed, which means if 2 mol of SO3 interact, then 2 mol of H₂O is formed.

Let’s find the molar mass:

M (H₂O) = 18 g / mol, so the mass of 2 mol of water is:

m (H₂O) = 2 mol * 18 g / mol = 36 grams.

Answer: 36 grams of water.