A35.0-ml sample of 1.00 m kbr and a 60.0-ml sample of 0.600 m kbr are mixed. the solution is then heated to evaporate water until the total volume is 50.0 ml. how many grams of silver nitrate are required to precipitate out silver bromide in the final solution?

The number of moles in the resulting solution can be calculated by adding together the number of moles from the individual solution. That is calculated as follows,
   
     total number of moles = (35 mL)(1 L/1000 mL)(1 mole/L) + (60 mL)(1 L/1000 mL)(0.6 moles / L) = 0.071 moles

The precipitation reaction is,

      KBr + AgNO3 --> AgBr + KNO3

From the equation, each mole of the KBr will need one mole of AgNO3. Hence, 0.071 mole of AgNO3 will also be needed. Calculating for the mass,
  
          mass of AgNO3 = (0.071 moles)(169.87 g / mol) = 12.06 g

ANSWER: 12.06 grams

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Answer is: the equilibrium  shifts toward the reactants, to the left side of chemical reaction.it is because lead(ii) nitrate dissociates in water: pb(no₃)₂ → pb²⁺(aq) + 2no₃⁻(aq).according to lechatelier's principle, the position of equilibrium moves to counteract thechange (more lead cations, common ion), the position of equilibrium will move so that the concentration ofproducts decrease and concentration of reactants of chemical reaction increase.read more on -