Calculate the mass of PbCl2 formed when an excess of 0.100 M solution of NaCl is added to 0.4L of 0.7 M Pb(NO3)2.

Pb(NO3)2 + 2 NaCl --> PbCl2 + 2 NaNO3

mols Pb(NO3)2 = M x L = approx 0.28

mols PbCl2 formed = mols Pb(NO3)2 (see the 1:1 ratio for Pb(NO3)2/PbCl2
Then g PbCl2 = mols PbCl2 x molar mass PbCl2.

To calculate the mass of PbCl2 formed, we need to use the information given regarding the molarity and volume of the solutions.

1. Calculate the number of moles of Pb(NO3)2:
To do this, we can use the formula: moles = molarity × volume (in liters)
moles(Pb(NO3)2) = 0.7 M × 0.4 L = 0.28 moles

2. Determine the limiting reactant:
To find the limiting reactant, we need to compare the stoichiometric ratios of the two reactants, Pb(NO3)2 and NaCl. According to the balanced equation, 1 mole of Pb(NO3)2 reacts with 2 moles of NaCl to form 1 mole of PbCl2. Therefore, we need 2 times the number of moles of NaCl compared to Pb(NO3)2 for a complete reaction.
moles(NaCl) needed = 2 × moles(Pb(NO3)2) = 2 × 0.28 moles = 0.56 moles

Since an excess of 0.100 M NaCl solution is used, we have more than enough NaCl to react completely with 0.28 moles of Pb(NO3)2. Therefore, Pb(NO3)2 is the limiting reactant.

3. Calculate the number of moles of PbCl2 formed:
Since 1 mole of Pb(NO3)2 reacts to form 1 mole of PbCl2, the number of moles of PbCl2 formed is also 0.28 moles.

4. Calculate the molar mass of PbCl2:
Pb has an atomic mass of 207.2 g/mol, and Cl has an atomic mass of 35.45 g/mol.
The molar mass of PbCl2 is 207.2 g/mol + (2 × 35.45 g/mol) = 278.1 g/mol.

5. Calculate the mass of PbCl2 formed:
mass(PbCl2) = moles(PbCl2) × molar mass(PbCl2)
mass(PbCl2) = 0.28 moles × 278.1 g/mol = 77.60 grams.

Therefore, the mass of PbCl2 formed when an excess of 0.100 M NaCl is added to 0.4 L of 0.7 M Pb(NO3)2 is 77.60 grams.