A double replacement reaction occurs between two solutions of lead (1) nitrate and potassium bromide, Write a balanced equation for this reaction-identifying the product that will precipitate, and the product that will remain in solution. If this reaction starts with 32.5 g lead (Il) nitrate and 38.75 g potassium bromide, how many grams of the precipitate will be produced? Remember to use the limiting reactant to calculate the amount of precipitate formed.
The balanced equation for the reaction between lead (II) nitrate (Pb(NO3)2) and potassium bromide (KBr) is:
Pb(NO3)2 + 2KBr -> PbBr2 + 2KNO3
In this reaction, the precipitate formed is lead (II) bromide (PbBr2). Potassium nitrate (KNO3) will remain in solution.
To determine the limiting reactant, we must first calculate the molar masses of the compounds:
Lead (II) nitrate (Pb(NO3)2): 207.2 g/mol + 2(14.01 g/mol + 3(16.00 g/mol)) = 331.2 g/mol
Potassium bromide (KBr): 39.10 g/mol + 79.90 g/mol = 119.0 g/mol
Next, calculate the number of moles of each reactant:
Number of moles of Pb(NO3)2 = 32.5 g / 331.2 g/mol = 0.098 mol
Number of moles of KBr = 38.75 g / 119.0 g/mol = 0.325 mol
Since the reaction requires 1 mole of Pb(NO3)2 to react with 2 moles of KBr to produce 1 mole of PbBr2, Pb(NO3)2 is the limiting reactant.
Now, calculate the amount of lead (II) bromide precipitate produced:
Number of moles of PbBr2 = 0.098 mol Pb(NO3)2 x (1 mol PbBr2 / 1 mol Pb(NO3)2) = 0.098 mol PbBr2
Mass of PbBr2 = 0.098 mol PbBr2 x 206.2 g/mol = 20.2 g
Therefore, 20.2 grams of lead (II) bromide will be produced in the reaction.