Question

Вычислите массу и объем кислорода , образовавшегося при разложении 72 г воды .

Answer 1

2Н2О = 2Н2 + О2.

М (Н2О) = 1 * 2 + 16 = 18 г/моль.
М (Н2) = 1 * 2 = 2 г/моль.
Составьте пропорцию.
72 г воды - х г водорода.
2 * 18 г/моль воды - 2 * 2 г/моль водорода.
Х = 72 * 2 * 2 : (2 * 18) = 8 г.
Далее определите массу кислорода.
М (О2) = 2 * 16 = 32 г/моль.
72 г воды - у г кислорода.
2 * 18 г/моль воды - 32 г/моль кислорода.
у = 72 * 32 : (2 * 18) = 64 г.
Ответ: m (H2) = 8 г; m (O2) = 64 г.

Answer 2

Calculating the Mass and Volume of Oxygen Produced from the Decomposition of 72g of Water

To calculate the mass and volume of oxygen produced from the decomposition of 72g of water, we can use the chemical equation for the decomposition of water into its constituent elements, hydrogen and oxygen.

The balanced chemical equation for the decomposition of water is: 2H2O(l) → 2H2(g) + O2(g)

From this equation, we can see that 2 moles of water decompose to produce 1 mole of oxygen gas.

Calculating the Mass of Oxygen Produced

To calculate the mass of oxygen produced, we can use the molar mass of water and the molar mass of oxygen.

The molar mass of water (H2O) is approximately 18.015 g/mol. The molar mass of oxygen (O2) is approximately 32.00 g/mol.

Given that the mass of water decomposed is 72g, we can calculate the mass of oxygen produced using the molar ratio from the balanced chemical equation.

Using the molar ratio, we find: - Moles of water decomposed = 72g / 18.015 g/mol = 3.994 moles - Moles of oxygen produced = 3.994 moles / 2 = 1.997 moles

The mass of oxygen produced can be calculated as: - Mass of oxygen = moles of oxygen * molar mass of oxygen - Mass of oxygen = 1.997 moles * 32.00 g/mol = 63.904g

Therefore, the mass of oxygen produced from the decomposition of 72g of water is approximately 63.904g.

Calculating the Volume of Oxygen Produced

To calculate the volume of oxygen produced, we can use the ideal gas law, which relates the volume of a gas to its amount, pressure, and temperature.

The ideal gas law is given by the equation: PV = nRT

Where: - P = pressure - V = volume - n = number of moles of gas - R = ideal gas constant - T = temperature

Given that the temperature and pressure are not provided, we can assume standard temperature and pressure (STP) for the calculation. At STP, the pressure is 1 atm and the temperature is 0°C (273.15 K).

Using STP, the volume of 1 mole of any gas is 22.4 liters.

Using the molar amount of oxygen produced (1.997 moles), we can calculate the volume of oxygen at STP: - Volume of oxygen = moles of oxygen * 22.4 L/mol - Volume of oxygen = 1.997 moles * 22.4 L/mol = 44.7488 L

Therefore, the volume of oxygen produced from the decomposition of 72g of water at STP is approximately 44.7488 liters.