Question

Fill in the Blanks When potassium iodide reacts with lead(II) nitrate, a yellow precipitate of lead iodide (PbI) and potassium nitrate are produced. Consider this reaction to answer the questions below: When properly balanced the coefficient in front of potassium iodide is the coefficient in front of lead(1) nitrate is , the coefficient in front of lead iodide is and the coefficient in front of potassium nitrate is If 0.78 g of lead() iodide was produced and you had an excess of potassium iodide, you must have started with a minimum of grams of lead(II) nitrate.

Answer 1

potassium Toclide reacts will lead (1) Nitrate to form Yellow preuptate of Leod Todide (PbI₂) Quel potassium Nitrate. - pbi, + k Noz KI + Pb(NO3), Let us balance tirar Nitrate (NO3) radical, KI + Pb(NO3), ) PbIz + 2 KNO3 . Let us belauce Iodine atais, 2KI+ Pb(NO3)2 -> DbI, () + 2K NOZ Yellow precipitate This is balanced equation, properly balanced, Coefficient introut of potassium Iocide is. [2] the coefficient of intraut of lead (D) Nitrate is a une coetticient of lead Iodice iss T and Coefficient of potassium Nitrate is: [2]. Given, 0.28 oot Lead (D) Jocide produced, Potassium Jodide is excess reagent the It is left after reaction. pilote Lead is limiting reagentice it is totally consumed in ele reaction Produced from Imole of DbI, Imole of Lead Netrate.

Molav mass of DbI₂ = 1x atauic mass of '86 + 2x Atauit mass of Iodine = 1*207+ 2x127 = 200+ 254 1. = 4618/un Molar mass of Pb(NO₃) = 1x Alamic mass of Db+ 2 маяк о+ 'N'+ 6xаlаша моях он ох9gец - 1x 207 + 2x1476*16 = 207+28+96 = 331 g/mol. To produce 461 9 of pbiz - 331 g of Db(NO3)2 produced, to produce 0.78 3 of Lead dodide - ? ? mase of Pb(NO3)z needed = 0.78*331 461 -0.56 g i minimum 0.56 got Pb(NO3)2 needed to produce 0.78 3 of Lead Jodide,