Hello I need help on caculating this problem that I got stuck on, and couldn\'t

Question

Hello I need help on caculating this problem that I got stuck on, and couldn't find the exact answer for this, can someone help me solve this? Thank you. Propane will react with molecular bromine to make 2,2-dibromopropane and hydrogen bromide. In this reaction, 6.21 grams of propane react with 13.1 mL of molecular bromine. 1. How many moles of propane are present before the chemicals react? 2. How many moles of molecular bromine are present before the chemicals react? 3. Which chemical is the limiting reagent? Enter a 1 for propane or enter a 2 for bromine. 4. How many grams of the excess reagent will be left over after the reaction? 5. How much 2,2-dibromopropane would be produced from this reaction? 6. If 22.8 grams of 2,2-dibromopropane were recovered from the reaction, what would the % yield be? 7. How many milliliters of 2,2-dibromopropane were recovered from this reaction? 8. In this reaction, the 2,2-dibromopropane is your desired product and the HBr is waste. What is the "atom economy" percent of this reaction?

Explanation / Answer

The reaction is:

C3H8 + Br2 --> CH3CBr2CH3 + 2HBr

1)1 mole of propane = 44 gms

      6.21gm of Propane = 6.21/44 = 0.141 moles
2)13.1 ml of Bromine = 0.255 moles
3)Propane is limiting reagent - 1
4) (0.255-0.141) moles of exces reagent left = 0.114*79.904*2 (Br atomic weight = 79.904 gm/mol)                                                                 =     18.2 gm
5) weight of 2,2-dibromopropane per mole = 203.808 gm

    0.141 mole is formed from reaction = 203.808*0.141 = 28.74 gm
6) 22.8 gm is the yield in the place of 28.74 gm

     % efficiency of reation = 22.8*100/28.74 = 79.33%
7)12.8 ml
8)55.49%

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