SHARE
TWEET

Untitled

a guest Apr 21st, 2019 80 Never
Not a member of Pastebin yet? Sign Up, it unlocks many cool features!
  1. molar mass of C4H10 = 58.1 g/mol
  2. molar mass of O2 = 32 g/mol
  3. number of mol = (given mass)/(molar mass)
  4. number of mol of C4H10 = 8.16/58.1
  5. = 0.141 mol
  6. number of mol of O2 = 15.7/32
  7. = 0.491 mol
  8.  
  9. reaction taking place
  10. 2C4H10 + 13O2 --> 8CO2 + 10H2O
  11. according to reaction
  12. 13 mol of O2 required 2 mol of C4H10
  13. 1 mol of O2 required 2/13 mol of C4H10
  14. 0.491 mol of O2 required (2/13)*0.491 mol of C4H10
  15. 0.491 mol of O2 required 0.0755 mol of C4H10
  16.  
  17. but we have only 0.141 mol of C4H10
  18. so, C4H10 is in excess
  19.  
  20. again,
  21. 13 mol of O2 give 8 mol of CO2
  22. 1 mol of O2 give 8/13 mol of CO2
  23. 0.491 mol of O2 give (8/13)*0.491 mol of CO2
  24. 0.491 mol of O2 give 0.302 mol of CO2
  25. so,
  26. number of mol of CO2 formed = 0.302 mol
  27.  
  28. moalar mass of CO2 = 44.0 g/mol
  29. mass of CO2 = (number of mol of CO2 formed)*(moalar mass of CO2)
  30. = 0.302*44.0
  31. = 13.3 g
  32.  
  33. formula for limiting reagent is O2
  34.  
  35. number of mol of C4H10 remains = (number of mol of C4H10) - (number of mol of C4H10 reacted)
  36. = (0.141 - 0.0755)
  37. = 0.0655 mol
  38.  
  39. mass of C4H10remains = (number of mol of C4H10 remains)*(molar mass of C4H10)
  40. = 0.0655*58.1
  41. = 3.81 g
RAW Paste Data
We use cookies for various purposes including analytics. By continuing to use Pastebin, you agree to our use of cookies as described in the Cookies Policy. OK, I Understand
 
Top