2K + Br2-> 2KBr
51.1 Br2 is reacted, and 74.1 KBr is produced. What is the precent yeild for the reaction?
Please explain how thus is to be anwsered
Chem
Moderators: Xen, expert, ChenBeier
Ok, since you are not given the amount of K used you can assume that it is in excess.
-First calculate the number of moles of Br2 (51.1g/159.8 = .31977 mol)
-Now when you look at the equation you will see that for every mol of Br2 you get 2mol of KBr, so you multiply molBr2 by two to get:
(.31977 * 2 = .6355 mol KBr)
-This number is the expected number of moles, but you want grams so you multiply by the molar mass of KBr (.6355 mol * 119g = 76.1g)
-So, you expected 76.1g and got 74.1g. To find the yield, divide what you got by what you expected:
74.1/76.1 *100% = 97.4% yield
Hope that helped
-First calculate the number of moles of Br2 (51.1g/159.8 = .31977 mol)
-Now when you look at the equation you will see that for every mol of Br2 you get 2mol of KBr, so you multiply molBr2 by two to get:
(.31977 * 2 = .6355 mol KBr)
-This number is the expected number of moles, but you want grams so you multiply by the molar mass of KBr (.6355 mol * 119g = 76.1g)
-So, you expected 76.1g and got 74.1g. To find the yield, divide what you got by what you expected:
74.1/76.1 *100% = 97.4% yield
Hope that helped