Question

2Al + 3ZnCl2 —> 3Zn + 2AlCl3

What mass of aluminum chloride will be produced if 22.6 grams of zinc chloride are used up in the reaction?

Answer 1

14.7 grams of aluminum chloride (AlCl₃) will be produced.

Step-by-step explanation:

In the reaction 2Al + 3ZnCl₂ → 3Zn + 2AlCl₃, for every 2 moles of aluminum, 2 moles of aluminum chloride (AlCl₃) are produced. So, if we know the moles of zinc chloride (ZnCl₂) used, we can calculate the moles of aluminum chloride produced by dividing by the stoichiometric coefficient of ZnCl₂ (3) and then multiplying by the stoichiometric coefficient of AlCl₃ (2).

We know that 22.6 grams of zinc chloride are used up in the reaction. To find the moles of zinc chloride, we can use the molar mass of zinc chloride, which is 136.29 g/mol.

22.6 g ZnCl2 / 136.29 g/mol ZnCl₂ = 0.166 mol ZnCl₂

So the moles of aluminum chloride produced would be 0.166 mol ZnCl₂ * 2/3 = 0.111 mol AlCl₃.

To find the mass of aluminum chloride produced, we just need to multiply the moles of AlCl₃ by its molar mass which is:

0.111 mol AlCl₃ * 133.34 g/mol AlCl₃ = 14.7 g AlCl₃

So, if 22.6 grams of zinc chloride are used up in the reaction, 14.7 grams of aluminum chloride will be produced.

Answer 2

To find the mass of aluminum chloride produced from 22.6 grams of zinc chloride, the amount in grams is converted to moles using the molar mass of ZnCl2, followed by applying the stoichiometric relationship from the balanced equation to find moles of AlCl3, which are then converted back to grams, resulting in 14.74 grams of aluminum chloride produced.

Step-by-step explanation:

The student's question asks what mass of aluminum chloride will be produced if 22.6 grams of zinc chloride are used up in the reaction 2Al + 3ZnCl2 → 3Zn + 2AlCl3. To solve this problem, we first need to find the molar mass of zinc chloride (ZnCl2) which is approximately 136.30 g/mol.

We then convert 22.6 grams of ZnCl2 to moles:

• 22.6 grams ZnCl2 × (1 mol ZnCl2 / 136.30 grams ZnCl2) = 0.1658 moles ZnCl2

Using the stoichiometry of the balanced equation, for every 3 moles of ZnCl2 that react, 2 moles of AlCl3 are produced. Thus, the moles of AlCl3 produced are calculated as follows:

• 0.1658 moles ZnCl2 × (2 mol AlCl3 / 3 mol ZnCl2) = 0.1105 moles AlCl3

The molar mass of aluminum chloride (AlCl3) is approximately 133.34 g/mol, so the mass of AlCl3 produced is:

• 0.1105 moles AlCl3 × (133.34 g AlCl3 / 1 mol AlCl3) = 14.74 grams AlCl3

Therefore, if 22.6 grams of zinc chloride are used up in the reaction, 14.74 grams of aluminum chloride will be produced.