Question

1. A 75.0-mL sample of 0.200 M Lead(II) nitrate, Pb(NO3)2, is reacted with 75.0 mL of 0.450 M KI solution and the following precipitation reaction occurs.

Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)

(a) Determine the limiting reactant. (b) How many grams of PbI2 will be formed if the yield is 100%? (c) What is the percent yield if 6.45 g of PbI2 were obtained?

(2) K2Cr2O7(aq) + FeCl2(aq) → CrCl3(aq) + Fe(NO3)3(aq)

(a) Determine the net ionic reactions and (b) balance the following equation for a redox reaction in acidic solution using the half-equation method. (c) Indicate the oxidizing agent and the reducing agent. (d) Identify the element being oxidized and the element being reduced.

3. Balance the following net ionic equation for a redox reaction in basic solution using the half-equation method. Indicate the oxidizing agent and the reducing agent. Identify the element being oxidized and the element being reduced.

Cr(OH)3(s) + H2O2(aq) → CrO42-(aq) + H2O(l)

Answer 1

Given: We know moles Mobodity a aume 9 Moles of Pb(NO), = 0.2008 0.075 = 0.015 moles Moles of KI = 0.450 X 0.075 = 0.03375 males Balanced Chemical reaction is given as; Pb/No. +2KI Pb I t 2 RNO 3 0.015 moles 0.033.75moles- Imole Pb (nog)e reacts with 2 moles of KI So 0.015 mole Pb(NO3) will react with 0.030 maly LOL KI but Ki is present in larger amount, So all Pb (NO2) will consumed in the reaction and It will be the limiting reagent. b) moles of Ph Iz - Moles of PINO; } formed consumed - E00015 moles so mass of Pb I (gm) = moles x helar mass - = 0.015 446156-915m C) Given: Actual yield = 6.459m Il percent yield • Actual yield (am) X 100 - Theoretical yield (am) © 6:45X100_E93e27 ofe 6-915