Question

Какой объём хлора требуется для полного сжигания 15 г алюминия? Какова масса ALCI3,который образуется? 2AL+3CI2=2ALCI3

Answer 1

2Al+3Cl₂=2AlCl₃
n(Al)=15/27=0,55 моль
$\frac{0,55}{2} = \frac{x}{3}$
x=n(Cl₂)=0,825 моль
V(Cl₂)=0,825·22,4=18,48 дм³
$\frac{0,55}{2} = \frac{y}{2}$
y=n(AlCl₃)=0,55 моль
m(AlCl₃)=0,55·133,5=73,425г.

Answer 2

Calculation of Chlorine Volume and Mass of AlCl3

To calculate the volume of chlorine required for the complete combustion of 15g of aluminum and the mass of AlCl3 formed, we can use the balanced chemical equation for the reaction:

2Al + 3Cl2 → 2AlCl3

First, let's calculate the amount of moles of aluminum (Al) in 15g:

The molar mass of aluminum (Al) is 27 g/mol. Therefore, the number of moles of aluminum can be calculated as:

Number of moles = Mass / Molar mass = 15g / 27 g/mol

Now, we can use the stoichiometry of the balanced chemical equation to determine the amount of chlorine (Cl2) required and the mass of aluminum chloride (AlCl3) formed.

According to the balanced chemical equation, 2 moles of aluminum react with 3 moles of chlorine to produce 2 moles of aluminum chloride.

Using this information, we can calculate the amount of chlorine required and the mass of aluminum chloride formed.

Calculation Steps:

1. Calculate the moles of aluminum. 2. Use the mole ratio from the balanced chemical equation to find the moles of chlorine required. 3. Convert the moles of aluminum chloride to grams.

Let's proceed with the calculations.