Ask Question
29 December, 18:27

Determine the limiting reactant. 2al (s) + 3br2 (g) →2albr3 (s)

+5
Answers (1)
  1. 29 December, 18:40
    0
    Missing question: For each of the following reactions, 25.0 g of each reactant is present initially.

    m (Al) = 25 g.

    n (Al) = m (Al) : M (Al).

    n (Al) = 25 g : 27 g/mol.

    n (Al) = 0,926 mol.

    m (Br₂) = 25 g.

    n (Br₂) = 25 g : 160 g/mol.

    n (Br₂) = 0,156 mol.

    Bromine is the limiting reactant, because it has less amount of substance.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Determine the limiting reactant. 2al (s) + 3br2 (g) →2albr3 (s) ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers