SHARE

TWEET

# Untitled

a guest
Apr 21st, 2019
81
Never

**Not a member of Pastebin yet?**

**, it unlocks many cool features!**

__Sign Up__- molar mass of C4H10 = 58.1 g/mol
- molar mass of O2 = 32 g/mol
- number of mol = (given mass)/(molar mass)
- number of mol of C4H10 = 8.16/58.1
- = 0.141 mol
- number of mol of O2 = 15.7/32
- = 0.491 mol
- reaction taking place
- 2C4H10 + 13O2 --> 8CO2 + 10H2O
- according to reaction
- 13 mol of O2 required 2 mol of C4H10
- 1 mol of O2 required 2/13 mol of C4H10
- 0.491 mol of O2 required (2/13)*0.491 mol of C4H10
- 0.491 mol of O2 required 0.0755 mol of C4H10
- but we have only 0.141 mol of C4H10
- so, C4H10 is in excess
- again,
- 13 mol of O2 give 8 mol of CO2
- 1 mol of O2 give 8/13 mol of CO2
- 0.491 mol of O2 give (8/13)*0.491 mol of CO2
- 0.491 mol of O2 give 0.302 mol of CO2
- so,
- number of mol of CO2 formed = 0.302 mol
- moalar mass of CO2 = 44.0 g/mol
- mass of CO2 = (number of mol of CO2 formed)*(moalar mass of CO2)
- = 0.302*44.0
- = 13.3 g
- formula for limiting reagent is O2
- number of mol of C4H10 remains = (number of mol of C4H10) - (number of mol of C4H10 reacted)
- = (0.141 - 0.0755)
- = 0.0655 mol
- mass of C4H10remains = (number of mol of C4H10 remains)*(molar mass of C4H10)
- = 0.0655*58.1
- = 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.