The molecular equation is
6 I2 + 12 NaOH = 2 NaIO3 + 10 NaI + 6 H2O
and the net ionic is
6 I2 + 12 OH- = 2 IO3- + 10 I- + 6 H2O
what is the balanced equation for Rubidium metal reacting with halogen iodine
2Al + 3I2 -> 2AlI3
Calcium + Iodine --> Calcium iodide1 Ca + 1 I2 --Δ--> 1 CaI2
Cl2 + 2NaI --> 2NaCl + I2
3IBr + 4NH3 --> NI3 + 3NH4Br
what is the balanced equation for Rubidium metal reacting with halogen iodine
2Al + 3I2 -> 2AlI3
Calcium + Iodine --> Calcium iodide1 Ca + 1 I2 --Δ--> 1 CaI2
6NaOH + 3I2 = 5NaI + NaIO3 + 3H2O Six moles of sodium hydroxide and three moles of diatomic iodine yield five moles of sodium iodide, one mole of sodium iodate, and three moles of water. Cheers!
kelage a tu? mercoir
2Cs (s) + I2 (l) = 2CsI (s)
Balanced:H2 + I2 ----> 2 HI
It does so in 1:3 ratio; hence the formula AlI3.Aluminum iodine is another example of iodine
H2,g + I2,s --> 2HIg
3IBr + 4NH3 --> NI3 + 3NH4Br
Cl2 + 2NaI --> 2NaCl + I2
Aluminium + Iodine ----> Aluminium iodide2 Al + 3 I2 ----> 2 AlI3