Question

Сколько грамм дихлорометана можно получить при взаимодействии 67,2 л метана с 89,6 л хлора?

Answer 1

Ответ: это по химии?

Объяснение:

Answer 2

Calculating the Amount of Dichloromethane Produced

To calculate the amount of dichloromethane produced when 67.2 L of methane reacts with 89.6 L of chlorine, we can use the balanced chemical equation for the reaction between methane and chlorine to form dichloromethane and hydrogen chloride.

The balanced chemical equation for the reaction is: CH4 + 2Cl2 → CH2Cl2 + 2HCl

According to the stoichiometry of the balanced chemical equation, 1 mole of methane reacts with 2 moles of chlorine to produce 1 mole of dichloromethane.

Now, let's calculate the moles of methane and chlorine involved in the reaction.

Moles of Methane and Chlorine

The molar volume of a gas at standard temperature and pressure (STP) is 22.4 L/mol.

- Moles of methane = Volume of methane / Molar volume at STP - Moles of methane = 67.2 L / 22.4 L/mol = 3 moles - Moles of chlorine = Volume of chlorine / Molar volume at STP - Moles of chlorine = 89.6 L / 22.4 L/mol = 4 moles

Stoichiometry Calculation

According to the balanced chemical equation, 1 mole of methane reacts with 2 moles of chlorine to produce 1 mole of dichloromethane.

Since there are 3 moles of methane and 4 moles of chlorine, the limiting reactant is methane, and it will be completely consumed in the reaction.

Amount of Dichloromethane Produced

As 1 mole of methane produces 1 mole of dichloromethane, the amount of dichloromethane produced will be equal to the moles of methane, which is 3 moles.

Now, let's convert the moles of dichloromethane to grams.

Conversion to Grams

The molar mass of dichloromethane (CH2Cl2) is approximately 84.93 g/mol.

- Grams of dichloromethane = Moles of dichloromethane * Molar mass - Grams of dichloromethane = 3 moles * 84.93 g/mol = 254.79 grams

Therefore, the amount of dichloromethane produced when 67.2 L of methane reacts with 89.6 L of chlorine is approximately 254.79 grams.