Question

M Al=30г m HCl=30г m (Alcl3) =?

Answer 1

Ответ:

2 Al + 6 HCl → 2 AlCl3 + 3 H2

n(Al)=m/M

n(Al)=30/27=1.11 mol

n(Al)=n(AlCl3)

m(AlCl3)=n*M

m(AlCl3)=1.11*133=147.63g

Объяснение:

Answer 2

To find the molar mass of aluminum chloride ($AlCl_3$), you need to know the individual molar masses of aluminum ($Al$) and chlorine ($Cl$).

The molar mass of aluminum ($Al$) is approximately $27 \, \text{g/mol}$, and the molar mass of chlorine ($Cl$) is approximately $35.5 \, \text{g/mol}$.

In $AlCl_3$, you have one aluminum atom and three chlorine atoms. Therefore, the molar mass of $AlCl_3$ is calculated as follows:

$$\text{Molar mass of } AlCl_3 = 1 \times \text{Molar mass of } Al + 3 \times \text{Molar mass of } Cl$$

$$= 1 \times 27 \, \text{g/mol} + 3 \times 35.5 \, \text{g/mol}$$

$$= 27 \, \text{g/mol} + 106.5 \, \text{g/mol}$$

$$= 133.5 \, \text{g/mol}$$

Now, you have $30 \, \text{g}$ of $Al$ and $30 \, \text{g}$ of $HCl$. To find out how much $AlCl_3$ can be formed, you need to determine the limiting reactant.

First, find the moles of $Al$ and $HCl$:

$$\text{Moles of } Al = \frac{30 \, \text{g}}{27 \, \text{g/mol}} = \frac{30}{27} \, \text{mol} \approx 1.11 \, \text{mol}$$

$$\text{Moles of } HCl = \frac{30 \, \text{g}}{36.5 \, \text{g/mol}} = \frac{30}{36.5} \, \text{mol} \approx 0.82 \, \text{mol}$$

Since $HCl$ is the limiting reactant (it produces fewer moles of $AlCl_3$), you can use the stoichiometry of the balanced chemical equation to find the moles of $AlCl_3$ formed. The balanced chemical equation for the reaction between $Al$ and $HCl$ is:

$2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2$

From the balanced equation, $2 \, \text{moles of } Al$ react with $6 \, \text{moles of } HCl$ to produce $2 \, \text{moles of } AlCl_3$.

Since you have $0.82 \, \text{mol}$ of $HCl$, the moles of $AlCl_3$ formed would be:

$\text{Moles of } AlCl_3 = 2 \times 0.82 \, \text{mol} = 1.64 \, \text{mol}$

Finally, to find the mass of $AlCl_3$, use its molar mass:

$\text{Mass of } AlCl_3 = \text{Moles of } AlCl_3 \times \text{Molar mass of } AlCl_3$ $= 1.64 \times 133.5 \, \text{g/mol} \approx 218.04 \, \text{g}$

Therefore, the mass of $AlCl_3$ formed is approximately $218.04 \, \text{g}$.