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You need to have values for K, Br2 and 2KBr in order to find out the balance. For example, if you had the values as H2 + O2 = H20 the balanced chemical equation would be 10 H2 = H20.
Br2(l) + 2KI(aq) --> 2KBr(aq) + I2(s)
2KBr + Cl2 ----> 2KCl + Br2
Balanced:2 K + Br2 ----> 2 KBr
2KBr + 2H2O----->2KOH + Br2 + H2(g) this is the reaction in electrolysis of KBr in aqueous solution
Cl2 + 2KBr -----> Br2 + 2KCl
2K2MnO4 + Br2 ----> 2KMnO4 + 2KBr
2KCl + Br2 ---> Cl2 + 2KBr
The chemical reaction is:Br2 + 2 K = 2 KBr
Br2 + Zn ----> ZnBr2
Br2(aq) + 2KI(aq) ==> 2KBr(aq) + I2(aq)
2KI+Br2 ---->2KBr +I2
The balanced equation is MgI2 + Br2 >> MgBr2 + I2....
2 K + Br2 -> 2 Kbr
2Na + Br2 --> 2NaBr
H2 + Br2 = 2 HBr
Br2 + 2KI --> 2KBr + I2 Bromine oxidizes iodide to iodine and is left as bromide.
2Na + Br2 ----> 2NaBr
2 Na + Br2 --> 2 NaBr
Take ethene. Add Br2 to get the dibromo derivative. Then do a double dehydrohalogenation with Alcoholic KOH to get ethyne. CH2=CH2 + Br2 ---> CH2Br-CH2Br + 2KOH(Alcoholic)---> CH-=CH + 2KBr + 2H2O
Potassium reacts with bromine to form the salt known as potassium bromide. 2K + Br2 --> 2KBr
First write a balanced chemical equation: 2K + Br2 ---> 2KBR Find the limiting reactant by using the moles of each element and determining which one gives you the smallest number of moles of potassium bromide. 2.92 mol K (2 mol KBr/2 mol K)= 2.92 mol KBr 1.78 mol Br2 (2 mol KBR/1 mol Br2)=3.56 mol KBr potassium is your limiting reactant so the max. number of moles of KBr that can be produced is 2.92 mol of KBr
Ca + Br2 = CaBr2 doesn't need to be balanced.