1. (3 pts) Will a micture of calciumbromide and lithiumsulfate result in a preci

Question

1. (3 pts) Will a micture of calciumbromide and lithiumsulfate result in a precipitate? If so, ive the formula unit of the 2. (8 pts) Write balanced total lonic equations for the following mixtures. If no reaction happens, write NR O- tcopper (I)nitratet annnoniumphosphate S) A9) 0.200 M sodiuni suitate with 40.0 mL 6.150 M barium nitra Ca) ->2NAU,No32 (41) (a) How many moles of precipitate will be formed (if any)? wa.so 1) ple Ba (b) What is the concentration of barium ions after you mix the solution and wait for the reaction (if any) to finish. @o. O.m

Explanation / Answer

1. CaBr2(aq) + Li2SO4(aq) ---------> CaSO4(s) + 2LiBr(aq)
    precipitate is CaSo4
2. 3CuNO3(aq) + (NH4)3PO4(aq) ----------> Cu3PO4(s) + 3NH4NO3(aq)
   3Cu^+(aq) +3NO3^-(aq) + 3(NH4)^+(aq) +PO4^3-(aq) ----------> Cu3PO4(s) + 3NH4^+(aq)+3NO3^-(aq)
   removal of spectator ions to get net ionic equation
   3Cu^+(aq) +PO4^3-(aq) ----------> Cu3PO4(s) net ionic equation
3. Na2So4(aq) + Ba(NO3)2 (aq) ----------> 2NaNO3(aq) +BaSO4(s)
   no of moles of Na2SO4 = molariTy * volum in L
                          = 0.2*0.03 = 0.006 moles
   no of moles of Ba(NO3)2 = molariTy * volum in L
                            = 0.15*0.04 = 0.006 moles
1 moles of Na2SO4 react with Ba(NO3)2 to gives 1 mole of BaSo4
0.006 moles of Na2So4 react with Ba(NO3)2 to gives 0.006 moles of BaSo4
4. total volume of solution = 0.03 + 0.04 = 0.07 mole

conc of Ba^2+ = no of moles/volume of solution in L
                  = 0.006/0.07   = 0.086M

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