Question

Consider the reaction: Al2S3 + 6 HCl → 2 AlCl3 + 3 H2S The reaction has a 29.1% yield. How many grams of Al2S3 are needed to react with 30.0 moles of HCl?
a) The molar mass of Al2S3 is 150.16 grams/mole.
b) The molar mass of HCl is 36.46 grams/mole.
c) The molar mass of AlCl3 is 133.33 grams/mole.
d) The molar mass of H2S is 34.086 grams/mole.
1. 219 grams
2. 467 grams
3. 1.20 grams
4. 751 grams
5. 180. grams
6. 2580 grams

Answer 1

Answer:

The correct answer is 1. 219 grams this is arrived at by considering the actual yield of the reacting compounds

Explanation:

We are given the equation for the chemical reaction as

Al2S3 + 6 HCl → 2 AlCl3 + 3 H2S

From where we have 1 mole of Al2S3  reacting with 6 moles of HCl to form the products

Thus 30 moles of HCl will require (1/6)×30 or 5 moles of Al2S3  

However 1 mole of Al2S3 is 150.16 grams thus 5 moles = 5×150.16 = 750.8 grams ≅ 751 grams

Percentage Yield = Actual Yield /Theoretical yield × 100%

As the reaction has a 29.1% yield, it shows that only 29.1 percent of the expected  product is actually produced

Hence the amount of Al2S3  required = (29.1/100)×751 = 218.5 grams ≅ 219 grams