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

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