How much hydrogen will be released if an alloy consisting of 4.6 g of sodium and 3.9 g of potassium is placed in water?

The reaction of sodium and potassium with water takes place according to the following chemical reaction equation:

Me + H2O = MeOH + ½ H2;

Where Me is a metal: sodium or potassium.

In accordance with it, when 1 mole of metal interacts with 1 mole of water, 0.5 mole of hydrogen gas is released.

Let’s calculate the amount of a substance contained in 4.6 grams of sodium and 3.9 grams of potassium.

M Na = 23 grams / mol;

N Na = 4.6 / 23 = 0.2 mol;

M K = 39 grams / mol;

N K = 3.9 / 39 = 0.1 mol;

N Me = N Na + N K = 0.2 + 0.1 = 0.3 mol;

During the reaction, 0.3 / 2 = 0.15 mol of hydrogen will be released.

Let’s calculate its volume.

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

The volume of hydrogen will be equal to:

V H2 = 0.15 x 22.4 = 3.36 liters;