Question

Помогите очень срочно решить задачу с оформлением какой объем хлора необходим для взаимодействия с 92 г натрия

Answer 1

Дано:
m(Na)=92 г.
Vm=22.4 л.
Найти: V(Cl2)=?
Решение:
92г   х л                              Решаем через пропорцию:
2Na+Cl2  =  2NaCl ;            х л=92г*22.4л/46г=44.8 литров.
46г  22.4л
Ответ: 44.8 литров.

Answer 2

Calculating the Volume of Chlorine Required to React with 92g of Sodium

To calculate the volume of chlorine required to react with 92g of sodium, we can use the balanced chemical equation for the reaction between sodium and chlorine. The balanced chemical equation for the reaction is:

2Na + Cl2 → 2NaCl

According to the stoichiometry of the balanced chemical equation, 1 mole of chlorine reacts with 2 moles of sodium to produce 2 moles of sodium chloride. From the molar mass of sodium (22.99 g/mol) and chlorine (35.45 g/mol), we can calculate the amount of chlorine required to react with 92g of sodium.

Using the molar masses of sodium and chlorine, we can calculate the moles of sodium and then determine the moles of chlorine required for the reaction.

The molar mass of sodium (Na) is approximately 22.99 g/mol. The molar mass of chlorine (Cl2) is approximately 35.45 g/mol.

Calculation:

1. Calculate the moles of sodium: Moles of sodium = Mass of sodium / Molar mass of sodium Moles of sodium = 92g / 22.99 g/mol Moles of sodium ≈ 4 moles

2. Determine the moles of chlorine required: According to the balanced chemical equation, 1 mole of chlorine reacts with 2 moles of sodium. Moles of chlorine required = Moles of sodium / 2 Moles of chlorine required ≈ 4 moles / 2 Moles of chlorine required ≈ 2 moles

3. Convert moles of chlorine to volume: As the question asks for the volume of chlorine, we can use the ideal gas law to convert moles to volume at standard temperature and pressure (STP), which is 0°C and 1 atm pressure.

Using the ideal gas law: PV = nRT Where: P = pressure (1 atm) V = volume n = moles of chlorine R = ideal gas constant (0.0821 L·atm/mol·K) T = temperature in Kelvin (273 K at STP)

V = nRT / P V = 2 moles * 0.0821 L·atm/mol·K * 273 K / 1 atm V ≈ 44.946 L

Conclusion:

The volume of chlorine required to react with 92g of sodium is approximately 44.946 liters at standard temperature and pressure (STP).