Determine the volume of sulfur (VI) oxide formed by the interaction of 4 g of sulfur and 8 g of oxygen.

1.Let’s find the amount of sulfur.

n = m: M.

M (S) = 32 g / mol.

n = 4 g: 32 g / mol = 0.125 mol.

Let’s find the amount of oxygen.

M (O2) = 32 g / mol.

n (O2) = 8 g: 32 g / mol = 0.25 mol.

Sulfur is in short supply. We solve the problem by sulfur.

Let’s compose the reaction equation:

2S + 3O2 = 2SO3.

According to the reaction equation, 2 mol of sulfur accounts for 2 mol of sulfur oxide, that is, the substances are in quantitative ratios of 1: 1, which means their amount of substances will be the same.

n (S) = n (SO3) = 0.125 mol.

Find the volume of sulfur oxide.

V = Vn n, where Vn is the molar volume of gas, equal to 22.4 l / mol, and n is the amount of substance.

V = 22.4 L / mol × 0.125 mol = 2.8 L.

Answer: 2.8 liters.