Question

Найти объём кислорода полученного при разложении 30 грамм воды

Answer 1

Сначала находишь количество моль воды
30/18=1,6
Далее количество моль кислорода
2/1=2 и 2*1,6=3,2
Затем объем 
3,2*22,4=71,68

Answer 2

Calculation of the Volume of Oxygen Produced from the Decomposition of 30 grams of Water

To calculate the volume of oxygen produced from the decomposition of 30 grams of water, we need to use the molar mass of water and the stoichiometry of the reaction.

The chemical equation for the decomposition of water is:

2H2O -> 2H2 + O2

From this equation, we can see that for every 2 moles of water decomposed, 1 mole of oxygen is produced.

To calculate the moles of water in 30 grams, we need to divide the mass of water by its molar mass. The molar mass of water (H2O) is approximately 18 grams/mol.

Moles of water = mass of water / molar mass of water Moles of water = 30 grams / 18 grams/mol Moles of water = 1.67 mol

Since the stoichiometry of the reaction is 2 moles of water to 1 mole of oxygen, we can calculate the moles of oxygen produced:

Moles of oxygen = Moles of water / 2 Moles of oxygen = 1.67 mol / 2 Moles of oxygen = 0.835 mol

Now, to calculate the volume of oxygen produced, we need to use the ideal gas law, which states:

PV = nRT

Where: P = pressure (in atm) V = volume (in liters) n = number of moles R = ideal gas constant (0.0821 L·atm/(mol·K)) T = temperature (in Kelvin)

Assuming standard temperature and pressure (STP), which is 0 degrees Celsius (273.15 Kelvin) and 1 atm, we can calculate the volume of oxygen:

V = nRT / P V = 0.835 mol * 0.0821 L·atm/(mol·K) * 273.15 K / 1 atm V ≈ 18.2 liters

Therefore, the volume of oxygen produced from the decomposition of 30 grams of water is approximately 18.2 liters.

Please note that this calculation assumes ideal conditions and may not account for any deviations from ideal behavior.