what mass of lead (II) iodide (PbI2, mass= 461 amu)is predictated 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)?
Pb(NO3)2 + 2Kl==>PbI2 + 2KNO3
You have the balanced equation.
1. Convert 50.0 mL of 0.60 M Pb(NO3)2 to moles. M x L = moles.
2. Using the coefficients in the balanced equation, convert moles Pb(NO3)2 to moles PbI2.
3. Now convert moles PbI2 to grams. g = mole x molar mass.
To determine the mass of lead (II) iodide (PbI2) precipitated in this reaction, you need to first calculate the moles of lead (II) nitrate (Pb(NO3)2) used and then use the stoichiometry of the reaction to convert the moles of lead (II) nitrate to moles of lead (II) iodide. Finally, you can convert the moles of lead (II) iodide to grams using its molar mass.
Let's break down the steps:
Step 1: Calculate the moles of lead (II) nitrate used.
Given:
- Volume of lead (II) nitrate solution = 50.0 mL = 0.0500 L
- Concentration of lead (II) nitrate solution = 0.60 M
Use the formula: Moles = Concentration x Volume
Moles of Pb(NO3)2 = 0.60 M x 0.0500 L = 0.0300 moles
Step 2: Use the stoichiometry of the reaction to determine the moles of lead (II) iodide formed.
From the balanced equation:
1 mole of Pb(NO3)2 reacts with 1 mole of PbI2.
So, the moles of PbI2 formed will be equal to the moles of Pb(NO3)2 used.
Moles of PbI2 = 0.0300 moles
Step 3: Calculate the mass of lead (II) iodide.
Given:
- Molar mass of PbI2 = 461 amu
Use the formula: Mass = Moles x Molar mass
Mass of PbI2 = 0.0300 moles x 461 amu = 13.83 grams of PbI2 (rounded to two decimal places)
Therefore, approximately 13.83 grams of lead (II) iodide will be precipitated.