what mass of lead (ii) iodide (pbl2, mass=461 amu)is precipitated by the addition of an excess of potassium iodide (ki, mass=166 amu)to 50.0 ml of 0.60 m lead (ii) nitrate (pb(no3)2, mass=331.2 amu)?
To calculate the mass of lead (II) iodide precipitated, we need to use the chemical equation and the stoichiometry of the reaction.
The balanced chemical equation for the reaction is as follows:
Pb(NO3)2 + 2KI → PbI2 + 2KNO3
From the equation, we can see that 1 mole of lead (II) nitrate reacts with 2 moles of potassium iodide to form 1 mole of lead (II) iodide.
Step 1: Calculate the moles of lead (II) nitrate present.
To find the moles of lead (II) nitrate, we can use the given volume and concentration:
Moles = Volume (in Liters) x Concentration (in M)
Moles = 0.0500 L x 0.60 mol/L (concentration of lead (II) nitrate)
Step 2: Calculate the moles of lead (II) iodide formed.
Since the reaction is 1:1 between lead (II) nitrate and lead (II) iodide, the moles of lead (II) iodide formed will be equal to the moles of lead (II) nitrate.
Step 3: Calculate the mass of lead (II) iodide formed.
The molar mass of lead (II) iodide (PbI2) is given as 461 amu.
Mass = Moles x Molar Mass
Mass of PbI2 = Moles of PbI2 x Molar mass of PbI2
Therefore, mass of PbI2 = Moles of PbI2 x 461 g/mol
Now, let's do the calculations:
Step 1: Moles of Pb(NO3)2:
Moles = 0.0500 L x 0.60 mol/L
Moles = 0.030 mol (lead (II) nitrate)
Step 2: Moles of PbI2 formed:
Moles of PbI2 = 0.030 mol (since the reaction is 1:1)
Step 3: Mass of PbI2:
Mass of PbI2 = Moles of PbI2 x Molar mass of PbI2
Mass of PbI2 = 0.030 mol x 461 g/mol
Mass of PbI2 = 13.83 g
Therefore, approximately 13.83 grams of lead (II) iodide (PbI2) will be precipitated by the addition of an excess of potassium iodide to the given solution.