Question

помогите пожалуйста 8 класс!!!!!Вычислите объем кислорода, нужны для сжигания 192 грамм меди, если в ходе реакции получается оксид меди (II)​

Answer 1

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Answer 2

Calculating the Volume of Oxygen Required for the Combustion of Copper Oxide (II)

To calculate the volume of oxygen needed for the combustion of 192 grams of copper to form copper oxide (II), we can use the stoichiometry of the reaction between copper and oxygen to determine the amount of oxygen required.

The balanced chemical equation for the formation of copper oxide (II) from copper and oxygen is:

2Cu + O2 → 2CuO

From the balanced equation, we can see that 1 mole of oxygen reacts with 2 moles of copper to form 2 moles of copper oxide (II).

Calculation Steps:

1. Calculate the moles of copper: - Given mass of copper = 192 grams - molar mass of copper (Cu) = 63.55 g/mol - moles of copper = given mass / molar mass - moles of copper = 192 g / 63.55 g/mol

2. Use the mole ratio from the balanced equation to find moles of oxygen: - According to the balanced equation, 1 mole of oxygen reacts with 2 moles of copper. - moles of oxygen = moles of copper / 2

3. Convert moles of oxygen to volume at standard conditions: - 1 mole of any gas at standard temperature and pressure (STP) occupies 22.4 liters. - volume of oxygen = moles of oxygen * 22.4 L/mol

Calculation:

1. Moles of copper: - Moles of copper = 192 g / 63.55 g/mol = 3.02 moles.

2. Moles of oxygen: - According to the balanced equation, 1 mole of oxygen reacts with 2 moles of copper. - Moles of oxygen = 3.02 moles / 2 = 1.51 moles.

3. Volume of oxygen at STP: - Volume of oxygen = 1.51 moles * 22.4 L/mol = 33.824 liters.

Therefore, the volume of oxygen needed for the combustion of 192 grams of copper to form copper oxide (II) is approximately 33.824 liters at standard conditions.