Hydrogen sulfide with a volume of 5.6 liters was burned in 11.2 liters of oxygen. Determine the masses of the reaction products.

Oxidation of hydrogen sulfide with oxygen is described by the following chemical reaction equation:

H2S + 3/2 O2 = H2O + SO2;

Let’s calculate the chemical amount of hydrogen sulfide. For this purpose, we divide its volume by the volume of 1 mole of gas (filling 22.4 liters).

N H2S = 5.6 / 22.4 = 0.25 mol;

Let’s find the molar amount of oxygen. To do this, we divide its volume by the volume of 1 mole of gas (which is 22.4 liters).

N O2 = 11.2 / 22.4 = 0.5 mol;

0.25 mol of hydrogen sulfide will react with 0.25 x 3/2 = 0.375 mol of oxygen. This will synthesize 0.25 mol of water and 0.25 mol of sulfur dioxide.

Let’s determine their weight. For this purpose, we multiply the amount of the substance by its molar mass.

M H2O = 2 + 16 = 18 grams / mol; m H2O = 0.25 x 18 = 4.5 grams;

M SO2 = 32 + 16 x 2 = 64 grams / mol; m SO2 = 0.25 x 64 = 16 grams;