PbI2 + HNO3 = Pb(NO3)2 + I2 + NO + H2O |

3PbI₂ + 8HNO₃ 3Pb(NO₃)₂ + 3I₂ + 2NO + 4H₂O

Lead(III) iodide react with nitric acid to produce nitrate lead(III), iodine, dinitrogen trioxide and water. Nitric acid - 30% solution. The reaction takes place in a boiling solution.

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Lead(III) iodide react with nitric acid

3PbI₂ + 8HNO₃ 3Pb(NO₃)₂ + 3I₂ + 2NO + 4H₂O

Picture of reaction:

Сoding to search: 3 PbI2 + 8 HNO3 cnd [ temp ] = 3 PbNO32 + 3 I2 + 2 NO + 4 H2O

Add / Edited: 04.07.2015 /

Evaluation of information: 5.0 out of 5 / number of votes: 1

Lead(III) iodide react with nitric acid

3PbI2 + 8HNO3 3Pb(NO3)2 + 3I2 + 2NO + 4H2O

Picture of reaction:

Сoding to search: 3 PbI2 + 8 HNO3 cnd [ temp ] = 3 PbNO32 + 3 I2 + 2 NO + 4 H2O

Add / Edited: 04.07.2015 / document.write(dtm['day']+'.'+dtm['mon']+'.'+dtm['year'])

3PbI₂ + 8HNO₃ 3Pb(NO₃)₂ + 3I₂ + 2NO + 4H₂O

Add / Edited: 04.07.2015 /