How much hydrogen must react with tungsten oxide (6) if 18.4 g of tungsten is formed?

The reduction of tungsten oxide with hydrogen occurs in accordance with the following chemical reaction equation:

WO3 + 3H2 = W + 3H2O;

From 1 mole of metal oxide, 1 mole of metal is reduced. This requires 3 moles of hydrogen gas.

Let’s calculate the amount of substance in 18.4 grams of metallic tungsten.

Its molar mass is:

M W = 184 grams / mol;

The amount of the substance will be:

N W = 18.4 / 184 = 0.1 mol;

This requires 0.1 x 3 = 0.3 mol of hydrogen.

Let’s calculate its volume.

One mole of gas under normal conditions fills a volume of 22.4 liters.

The volume of hydrogen will be:

V H2 = 0.3 x 22.4 = 6.72 liters;