Ask Question
24 March, 01:43

5. A compound with a formula of lead (Pb) and sulfur (S) was determined using the method in this experiment. A sample of Pb was weighed into a porcelain crucible and covered with finely powdered S. Then the crucible was covered and heated to allow the Pb and S to react. After additional heating, all the unreacted S was burned off. The crucible was then cooled, weighed, and finally showed the following dа ta: Mass of crucible and cover, 25.429 grams Mass of crucible, cover, and Pb, g 29.465 grams Mass of crucible, cover, and compound formed, g 30.107 grams What is the mass of S that reacted?

+3
Answers (1)
  1. 24 March, 03:32
    0
    The mass of S that reacted is 0.642g

    Explanation:

    In this case we have

    ⇒ mass of crucible and cover, 25.429g

    ⇒ mass of crucible, cover, and Pb, 29.465g

    ⇒ mass of crucible, cover and commass of crucible and coverpound formed, 30.107g

    The mass of Pb reacted = mass of crucible, cover, and Pb - mass of crucible and cover = 29.465 - 25.429 = 4.036g

    The mass of compound formed = mass of crucible, cover and commass of crucible and coverpound formed - mass of crucible and cover = 30.107 - 25.429g = 4.678g

    The mass of S reacted = Mass of compound formed - mass of Pb reacted = 4.678 - 4.036 = 0.642g

    The mass of S that reacted is 0.642g
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “5. A compound with a formula of lead (Pb) and sulfur (S) was determined using the method in this experiment. A sample of Pb was weighed ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers