Question

Aluminum chloride is named as if it is ionic but it is really molecular, with the formula Al2Cl6. It can be formed by direct reaction of the elements. 2Al(s) + 3Cl2(g) → AICIA) If 0.824 g of aluminum react with excess chlorine, how many grams of aluminum chloride can be obtained? (4.07 g) Calculate the grams of chlorine that react with 0.194 g of aluminum. (0.765 g) 1.2.b

Answer 1

1) 4.0744 grams of aluminum chloride can be obtained.

2) 0.7652 grams of chlorine that react with 0.194 g of aluminum.

Step-by-step explanation:

`2Al(s) + 3Cl_2(g)\rightarrow 2AICI_3`

1) Moles of aluminum :
`(0.824 g)/(27 g/mol)=0.03052 mol`

According to reaction, 2 moles of aluminum gives 2 moles of aluminium chloride.

Then 0.03052 mol of aluminum will give with :

`(2)/(2)* 0.03052 mol=0.03052 mol` of aluminium chloride.

Mass of 0.03052 moles of aluminium chloride :

`0.03052 mol* 133.5 g/mol=4.0744 g`

4.0744 grams of aluminum chloride can be obtained.

2) Moles of Aluminium :
`(0.194 g)/(27 g/mol)=0.007185 mol`

According to reaction, 2 moles of aluminum reacts with 3 moles of chlorine gas.

Then 0.007185 mol of aluminum will react with:

`(3)/(2)* 0.007185 mol=0.01077 mol` of chlorine gas.

Mass of 0.01077 mol moles of chlorine gas:

`0.01077 mol* 71 g/mol=0.7652 g`

0.7652 grams of chlorine that react with 0.194 g of aluminum.