Question

Aluminum metal reacts with hydrochloric acid to form aluminum chloride according to the reaction below. If 30.8 grams of aluminum react with 84 grams of HCl, how many grams of aluminum chloride can theoretically be formed? 2Al + 6HCl --> 2AlCl3 + 3H2 You Answered

Answer 1

Answer: The Theoretical yield of Aluminum Chloride = 152.19 g

Explanation:

In order to determine the theoretical yield of Aluminum Chloride we first determine the limiting reagent,

Al = 27g/mol

HCl = 1 + 35.5 = 36.5g/mol

number of moles= mass/molar mass

for Al:

number of moles of Al = 30.8/27 = 1.14 moles

number of moles of HCl = 84/36.5 = 2.30moles

Hence, Al is the limiting reagent

Therefore,

since from the balanced reaction,

2moles of Al react with HCl to give 2moles of Aluminum Chloride

Therefore 1.14moles Al will give 1.14moles of Aluminum Chloride.

molar mass of Aluminum Chloride = 27 + (35.5)*3 = 133.5g/mol

Theoretical yield of Aluminum Chloride = number of moles x molar mass = 1.14 x 133.5 = 152.19 g

Answer 2

Answer:

The amount in grams of aluminum chloride that can theoretically be formed from the 84 g of HCl and 30.8 g aluminium is 102.4 grams

Explanation:

Aluminum metal reacts with hydrochloric acid to form aluminum chloride according to the reaction below. If 30.8 grams of aluminum react with 84 grams of HCl, how many grams of aluminum chloride can theoretically be formed? 2Al + 6HCl --> 2AlCl3 + 3H2

Hydrochloric acid / Molar mass

36.46 g/mol

Aluminium chloride / Molar mass

133.34 g/mol

Aluminium / Atomic mass

26.981539

Hydrogen / Atomic mass

1.00784 u

Number of moles of aluminium = mass/molar mass = 30.8 g/ 26.98 g/mol

= 1.142 moles.

Number of moles of HCl = mass/molar mass = 84/36.46 = 2.303 moles

From the reaction, 2 moles of Al react with 6 moles of HCl, therefore, 1.142 moles of aluminium react with (1.142 × 6/2) moles or 3.425 moles of HCl. Since the calculated number of moles of required HCl is more than the available moles, then HCl is the limiting reagent.

6 moles of HCl react with 2 moles of aluminium, then 2.303 moles will react with (2/6×2.303) or 0.76796 moles of aluminium to produce 0.76796 moles of aluminum chloride

Therfore mass of aluminum chloride produced = (Number of moles of aluminium chloride produced) × (molar mass of aluminium chloride)

0.76796 moles × 133.34 g/mol = 102.4 g