Given the data: 1. Mass of test tube + MNO² = 21.5403g 21.5403g 2. Mass of the test tube + MnO² + KClO³ + KCl 22.5518g 3. Constant mass of test tube + reaction product 22.3514g a. Mass of O2 released: Answers by an expert of bartleyby: 1) 0.078 moles 2) 0.2004g b. Moles of O2 released: 1) 0.00243 moles 2) 0.0063 mol c. Moles of KClO3 required by the moles 1) 0.00163 moles d. Mass of KClO3 required: 1) 0.5116g g KClO3 d. Mass KClO3 & KCl mixture: 1) 1.0115g
I'm very confused. I didnt knew that 3 subtopics could be answered, so I uploaded one subtopic at the time. After gerting teh answers back, I uploaded the entire question and got the answer for the first 3 subtopics ( I'll call this A). However, a few of the single questions previously uploaded were answered and they are different from A. I'll upload the entire document again and please answer subtopics c, . If possible please confirm the write answer!
Given the data:
1. Mass of test tube + MNO² = 21.5403g 21.5403g
2. Mass of the test tube + MnO² + KClO³ + KCl 22.5518g
3. Constant mass of test tube + reaction product 22.3514g
a. Mass of O2 released:
Answers by an expert of bartleyby:
1) 0.078 moles
2) 0.2004g
b. Moles of O2 released:
1) 0.00243 moles
2) 0.0063 mol
c. Moles of KClO3 required by the moles
1) 0.00163 moles
d. Mass of KClO3 required:
1) 0.5116g g KClO3
d. Mass KClO3 & KCl mixture:
1) 1.0115g
e. Percent composition of KClO3 & KCl mixture
1) 50.56%
f. If there were 0.5011g KClO3 in the original mixture, what is the true percent of KClO3?
Trending now
This is a popular solution!
Step by step
Solved in 3 steps with 2 images