Question

4. Based on the activity series, what is the outcome (if any) of each of the following reactions? Where the reaction occurs, balance the chemical equation. If the reaction docs not occur state NR. (5 points) a. Mn(s) + NiCl(aq) → b. Cu(s) + Cr(CH,COO)3(aq) → c. Cr(s) + NiSO4(aq) → d. Pt(s) + HBr(aq) → e. H2(g) + CuCl(aq) → 5. A sample of 1.50 g of lead(II) nitrate is mixed with 125 mL of 0.100 M sodium sulfate solution. (6 points) a. Write the chemical equation for the reaction that occurs b. Which is the limiting reactant in the reaction? c. What is the theoretical yield of precipitate that forms? d. Extra Credit (3 points). What are the concentrations of all ions that remain in solution after the reaction is complete? Assume the volume of the final solution 125 mL.

Answer 1

70 Marsh Nicelag) → Mocy(at)t Al on Eletrochemical series Mn is above Ni em can replace NET from 99. Sunim I cw is becow or in (b) Cu(s) + Ofelzcov) 3 (99) NR W is below or in 7 the serves] the (Agholt 3 nison (29) Celson), + 3 Ni [or is above Ni in the - series) 20 (d) Pt (5) + ABB (11) HR [Pt is the low it in the series] - (@) W219) +esch/29) > 2H & Cues) [12 As above is in the serise] Eletrochemi.coy Series Elements are arranged on the basis of oxidation potential. Elements which is present top has more positive value of oxidation potential than an element of below position. So element on above position have more tendency to be oxidised ] 125 m 0.150m 5) Pb(NO3)2: Mo. Wt= 207+(14+ 16x3172 Mazsoy = 3318 = 1254 0.100 monal Maysoy 1.50g Pb(NO3) = 1.3 mol = 0.0045 my 1 = 125 x 0.17 16-3 mol Nasty = 0.0125 mol Nasou - Chemical equation: Pb(NO3)27 Mayson PbSO4 + 2NaNO3(aq) (b) Pb(NO3)2 present in less amount . So writing reagent is Pb(N63)2 - (c) So amount of ppt = Amount of amoting reagent "45 mol 7 Pbsog (Theorioileal yield) 1 (d) So after completion a reyh there will be 240.0045 my of Nanoz 10.0125-0.0045) mol J Mazsoy = 0.008 mal of Mazsoy. . Total no of moes of Nat= (2+0.0045) + (2x0.008) = 0.025 mer Total hoof moles of Noz = 2x0.5043= 0.009 & mores of Sop2- = 0.008 : -0.0045 mol of Posoy Totalno of mores of Sout- =

O [Mat)= 0,0235 1000 (m) = 0.216|| [ N03 )= 0.xlovo (m)> 0 .072(m) [sog2 -] = 0.0087 1000 (m) = 0.069 (m/