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?

Question

Chemistry

Carl Sutton

25 February, 11:50

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?

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Answers (1)

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Axel Bullock · Answer 1

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