Question

объемы CH4 и Cl2 которые израсходуется для получения 202 грамм хлорметана

Answer 1

CH4+Cl2= CH3Cl + HCl (над стрелкой квант света ашню или просто свет)

 

 

Answer 2

CH4+Cl2==(свет)==>CH3Cl+HCl
n(CH3Cl)=m/M=202/50.5=4 моль=n(CH4)
V(CH4)=n*Vm(молярный объем)=4*22.4= 89.6 л
Ответ: 89.6 л

Answer 3

Calculating the Volume of CH4 and Cl2 for the Production of 202 Grams of Chloromethane

To calculate the volumes of CH4 and Cl2 required to produce 202 grams of chloromethane, we can use the balanced chemical equation for the reaction between methane (CH4) and chlorine (Cl2) to produce chloromethane (CH3Cl) and hydrogen chloride (HCl).

The balanced chemical equation for the reaction is: CH4 + Cl2 → CH3Cl + HCl

To calculate the volumes of CH4 and Cl2, we need to use the stoichiometry of the reaction and the molar masses of the substances involved.

First, let's calculate the molar mass of chloromethane (CH3Cl) using the atomic masses of carbon (C), hydrogen (H), and chlorine (Cl).

The molar mass of CH3Cl is: M(CH3Cl) = (1 × M(C)) + (3 × M(H)) + M(Cl) M(C) = 12.01 g/mol (atomic mass of carbon) M(H) = 1.008 g/mol (atomic mass of hydrogen) M(Cl) = 35.45 g/mol (atomic mass of chlorine)

M(CH3Cl) = (1 × 12.01) + (3 × 1.008) + 35.45 M(CH3Cl) ≈ 50.49 g/mol

Now, we can use the molar mass of CH3Cl to calculate the number of moles of CH3Cl produced from 202 grams of CH3Cl.

The number of moles of CH3Cl is given by: n(CH3Cl) = Mass of CH3Cl / Molar mass of CH3Cl n(CH3Cl) = 202 g / 50.49 g/mol n(CH3Cl) ≈ 4 moles

Now, using the balanced chemical equation, we can determine the stoichiometric ratio of CH4 to CH3Cl.

The stoichiometric ratio of CH4 to CH3Cl is 1:1, which means 1 mole of CH4 produces 1 mole of CH3Cl.

Therefore, the number of moles of CH4 required is also approximately 4 moles.

Similarly, using the balanced chemical equation, we can determine the stoichiometric ratio of Cl2 to CH3Cl.

The stoichiometric ratio of Cl2 to CH3Cl is 1:1, which means 1 mole of Cl2 produces 1 mole of CH3Cl.

Therefore, the number of moles of Cl2 required is also approximately 4 moles.

To calculate the volumes of CH4 and Cl2 at standard temperature and pressure (STP), we can use the molar volume of a gas at STP, which is approximately 22.4 liters per mole.

So, the volumes of CH4 and Cl2 required at STP are both approximately 4 moles × 22.4 L/mole = 89.6 liters.

Therefore, to produce 202 grams of chloromethane, approximately 89.6 liters of CH4 and 89.6 liters of Cl2 are required.

[[7 #]]