Question

Помогите пожалуйста решить задачи))) 1. Какой объем хлора вступит в реакцию с порошком алюминия, если в результате реакции образуется 53,4 хлорида алюминия.2. При сгорании угля выделилось 176г углекислого газа. Какой объем кислорода израсходовался при этом?

Answer 1

1. 3Cl+Al=AlCl3
Cl= (67,2•53,4)/133,4=26,88

Answer 2

Calculating the Volume of Chlorine Reacting with Aluminum Powder

To calculate the volume of chlorine that will react with aluminum powder, we can use the ideal gas law equation, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles of the gas, R is the ideal gas constant, and T is the temperature.

First, we need to find the number of moles of chlorine that will react with the aluminum powder. We can then use this information to calculate the volume of chlorine.

Given: - Mass of aluminum chloride produced = 53.4 g - Equation for the reaction: 2Al + 3Cl2 → 2AlCl3

Calculating the Moles of Chlorine

Using the equation, we can determine the moles of aluminum chloride produced and then use stoichiometry to find the moles of chlorine involved in the reaction.

The molar mass of aluminum chloride (AlCl3) is 133.34 g/mol.

53.4 g of AlCl3 is produced, so the moles of AlCl3 can be calculated as: moles = mass / molar mass moles = 53.4 g / 133.34 g/mol moles ≈ 0.4 mol

According to the balanced chemical equation, 2 moles of aluminum chloride are produced from 3 moles of chlorine. Therefore, the moles of chlorine involved in the reaction are: moles of Cl2 = (3/2) * moles of AlCl3 moles of Cl2 = (3/2) * 0.4 mol moles of Cl2 = 0.6 mol

Calculating the Volume of Chlorine

Now that we have the moles of chlorine, we can use the ideal gas law to calculate the volume of chlorine at standard temperature and pressure (STP), which is 0°C and 1 atm.

The ideal gas constant, R, is 0.0821 L·atm/(K·mol), and the temperature, T, is 273 K (0°C in Kelvin).

Using the ideal gas law equation, we can solve for the volume, V: V = (nRT) / P V = (0.6 mol * 0.0821 L·atm/(K·mol) * 273 K) / 1 atm V ≈ 14.2 L

The volume of chlorine that will react with the aluminum powder is approximately 14.2 liters.

Calculating the Volume of Oxygen Consumed in Coal Combustion

To calculate the volume of oxygen consumed in the combustion of coal, we can use the stoichiometry of the reaction and the ideal gas law.

The balanced chemical equation for the combustion of coal is: C + O2 → CO2

Given: - Mass of carbon dioxide produced = 176 g - Molar mass of carbon dioxide (CO2) = 44.01 g/mol

Calculating the Moles of Carbon Dioxide

The moles of carbon dioxide produced can be calculated as: moles = mass / molar mass moles = 176 g / 44.01 g/mol moles ≈ 4 mol

According to the balanced chemical equation, 1 mole of carbon dioxide is produced from 1 mole of oxygen. Therefore, the moles of oxygen involved in the reaction are equal to the moles of carbon dioxide produced.

Calculating the Volume of Oxygen

Using the ideal gas law at STP, we can calculate the volume of oxygen: V = nRT / P V = 4 mol * 0.0821 L·atm/(K·mol) * 273 K / 1 atm V ≈ 36.1 L

The volume of oxygen consumed in the combustion of coal is approximately 36.1 liters.