Question

Aluminum chloride reacts with sodium sulfide to produce aluminum sulfide and sodium chloride. If one starts with 3.50 moles of aluminum chloride and excess sodium sulfide, how many moles of aluminum sulfide will be produced?

Answer 1

For 3.50 moles of aluminum chloride reacting with excess sodium sulfide, 1.75 moles of aluminum sulfide would be produced.

Step-by-step explanation:

Aluminum chloride reacts with sodium sulfide to produce aluminum sulfide and sodium chloride. Given that one starts with 3.50 moles of aluminum chloride and excess sodium sulfide, we must first write the balanced chemical equation for the reaction:

• 2 AlCl3 + 3 Na2S → Al2S3 + 6 NaCl

From the balanced equation, we see that 2 moles of AlCl3 produce 1 mole of Al2S3. So, if we have 3.50 moles of AlCl3, the moles of Al2S3 produced would be:

(3.50 moles AlCl3) × (1 mole Al2S3 / 2 moles AlCl3) = 1.75 moles of Al2S3

Therefore, for 3.50 moles of AlCl3, 1.75 moles of Al2S3 will be produced in the presence of excess sodium sulfide.