Homework SourseAtmospheric Pressure 7671 TORR Need Volume of gasmL Moles o
Atmospheric Pressure = 767.1 TORR
Need Volume of gas(mL), Moles of Oxygen gas, Moles of KClO3, Ratio: Moles of Oxygen gas:Moles of KClO3 Please show work so I can understand this :)! Thank you very much!
| Trial | 1 | 2 |
| Mass of Empty Test Tube | 18.51g | 18.53g |
| Mass of Test tube and KClO3 | 19.55g | 19.54g |
| Mass of KClO3 | 1.04g | 1.01g |
| Volume of water displaced(mL) | 370mL | 353mL |
| Temperature of water C | 25C | 25.5C |
| Mass of test tube and product | 19.1g | 19.08g |
| Mass of product | .59 | .55 |
Solution
Trial 1:
Mass of product = 19.55 gm - 19.1 gm = 0.45 gm (what you got in the table is not correct)
Moles of O2 gas, x = 0.45 gm / 32gm/mol = 0.0140625 mol
Moles of KClO3, y = 1.04 gm / 122.55 gm/mol = 8.48633*10-3 mol
Ratio = x : y = 0.0140625 / 8.48633*10-3 = 1.6571
Volume of gas = volume of water displaced = 370 mL
Trial 2 :
Mass of product = 19.54 gm - 19.08 gm = 0.46 gm
Moles of O2 gas = 0.46 gm / 32 gm/mol = 0.014375 mol
Moles of KClO3 = 1.01 gm / 122.55 gm/mol = 8.24153*10-3 mol
ratio = 0.014375 / 8.24153*10-3 = 1.7442
Volume of gas = 353 mL
