Question

Question 4 1.2 pts Fifty five (55.00) mL of 0.456 M aqueous RbOH solution is mixed with 45.00 mL of 0.302 M aqueous solution of Pb(NO3)2 The number of moles of RbOH available is The number of moles of Pb(NO3)2 available is The amount in grams of the precipitate formed is (MM (g/mole) RbOH = 102.478; Pb(OH)2 = 241.216; Pb(NO3)2=331.21; RbNO3=147.47) (Express your answer with 4 decimal figures or 4 prd's and do NOT include the units). (Hint: this problem is similar to textbook example 4.3 done in class)

Answer 1

solution 4 Reaction betweren Rb oH and Pb(NO3)2 Can be representea) RbNO₃ + Pb(OH)2 Rblon) + Pb(NO3) 1998 ( s Balance reaction is 2 Rb (or) of Pb(NO3) 2 2 RbNO3 + Pblond & laq) (5) (precipitate (29) (9) The number of moles of Rbol = 0.456 mol x 55.0 L 1000 = 0.02508 mol - 0.0251 mol (Ans) I The number of moles of Pb(NO₃) = 0.302 mol x 450 L wer of moes q Pb(NO3)2 - 1000 = 0.01359 mol. -0.0136 mol Ans since moles of Pb(NO3)2 is dess than Rbon is limiting seagout. So so. Pb(NO) Precipitate moles of pblon) formed = 0.0136 mol .: mass of pblow = mot of Pb (on ₂ x molar mats of Phong, -0.0136 X 241.216 28, 2805376 = 3.2805 g Ans)

Thus. The number of moles of Rb oh available is 10.07517 The number of moles of Pb (NO₂) available is (0.0136 ] The amount in grams of the precipitate formed 's [3-2805 ] Com e - Here before available means No. of moles ofe the reaction take place.