Calculate the mass of water that reacted 9.75 g of potassium. How much hydrogen was released during this?

Let’s execute the solution:

According to the condition of the problem, we write down the equation of the process:
m = 9.75 g Xg -?

2K + 2H2O = 2KOH + H2 – OBP, hydrogen and potassium hydroxide are released;

2.Calculation:

M (K) = 39.1 g / mol;

M (H2O) = 18 g / mol;

Y (K) = m / M = 9.75 / 39.1 = 0.25 mol;

Y (H2O) = 0.25 mol, since the amount of substances is equal to 2 mol.

We find the mass of H2O:
m (H2O) = Y * M = 0.25 * 18 = 4.5 g (deficient substance).

The calculation is made for the substance in deficiency.

Proportion:
0.25 mol (H2O) – X mol (H2);

-2 mol -1 mol hence, X mol (H2) = 0.25 * 1/2 = 0.125 mol.

Let’s determine the volume of the product:
V (H2) = 0.125 * 22.4 = 2.8 L

Answer: 2.8 liters of hydrogen was obtained.