Question

Объем хлора, необходимый для реакции присоединения с пропином оюъемом 11,2л?

Answer 1

CH3-C≡CH+Cl2-->CH3-CHCl=CHCl

U(C3H4)=V/V молярный=11,2л /22,4 л/моль=0,5 моль

из уравнения U(Cl2)=U(C3H4)=0,5 моль

V(Cl2)=U*V молярный=0,5 моль*22,4 л/моль=11,2 л

Answer 2

Calculating the Volume of Chlorine Required for the Reaction with 11.2 Liters of Propyne

To calculate the volume of chlorine required for the reaction with 11.2 liters of propyne, we can use the balanced chemical equation for the addition reaction between propyne (C3H4) and chlorine (Cl2):

C3H4 + Cl2 → C3H5Cl

The stoichiometry of the reaction tells us that 1 mole of propyne reacts with 1 mole of chlorine to produce 1 mole of 1-chloropropane. From this, we can calculate the volume of chlorine required.

Calculation

Given: - Volume of propyne (C3H4) = 11.2 liters

We need to find the volume of chlorine (Cl2) required for the reaction.

Using the ideal gas law, we can calculate the volume of chlorine required. The molar volume of an ideal gas at standard temperature and pressure (STP) is 22.4 liters/mol.

First, we need to convert the volume of propyne to moles using the ideal gas law:

n = PV/RT

Where: - P = pressure (at STP, P = 1 atm) - V = volume of propyne (11.2 liters) - R = ideal gas constant (0.0821 L·atm/mol·K) - T = temperature in Kelvin (at STP, T = 273 K)

Let's calculate the moles of propyne:

n(propyne) = (1 atm) * (11.2 L) / (0.0821 L·atm/mol·K * 273 K)

After finding the moles of propyne, we can use the stoichiometry of the reaction to determine the moles of chlorine required. Since the reaction is 1:1, the moles of chlorine required will be the same as the moles of propyne.

Finally, we can convert the moles of chlorine to volume using the molar volume of an ideal gas at STP:

Volume of chlorine = moles of chlorine * molar volume at STP

Let's calculate the volume of chlorine required based on the given information.