Question

excess KBr solution is added to 69.0 mL 14. How many grams of PbBr2 will precipitate when of 0.609 M Pb(NO); solution? Pb(NO)2 (aq)+ 2KB (aq) PbBr2(s)+2KNO,(aq) D) 15.4 g C) 13.4 g A) 12.4 g B) 1.5 g

Answer 1

Ans) Since, excess of KBr solution is used, then according to the given reaction ,Pb(NO₃)₂will act as the limiting reagent.

   Therefore, we firstly need to calculate the moles of Pb(NO₃)₂ reacting with excess of KBr.

              

→ Moles of Pb(NO3)2 = Volume * Concentration

                                                             

= 69 mL * (0.609 M)

                                                             

= 42.021 mmol = 42.021 * 10-3 mol

Now, from the balanced reaction, it is clear that 1 mol of Pb(NO₃)₂ forms 1 mol of PbBr₂ .

   
:. Moles of PbBr2 formed = 42.021 * 10-3

    Thus,
Mass of PbBr2 formed = Moles * Molar Mass

                                                              
= 42.021 * 10-3 mol * 367.01 g/mol

                                                              
= 15.42 g

   Hence, the correct option is d) 15.4 g .