Answer:
1.8 g of Pb(NO3)2
Explanation:
Find the moles of sodium chloride used by multiplying the molarity by the volume of sodium chloride.
Molarity = mol/L
Convert 23.5 mL to L.
23.5 mL x (1 L/1000 mL) = 0.0235 L
Multiply molarity by volume.
0.55 M = mol/0.0235 L
(0.55 M)(0.0235 L) = mol
mol = 0.012925
You have the moles of sodium chloride used so you can convert this to moels of lead (II) nitrate with stoichiometry. First, you need the balanced chemical equation.
Pb(NO3)2 + 2 NaCl -> PbCl2 + 2NaNO3
Convert 0.012925 mol NaCl with mole to mole ratio. In this case, it's 1:2.
0.012925 mol NaCl x (1 mol PbCl2/2 mol NaCl) = 0.0064625 mol PbCl2
Convert moles of PbCl2 to grams with molar mass.
0.0064625 mol PbCl2 x (278.10 g/1 mol) = 1.79722... g
Round to sig figs.
1.8 grams of PbCl2
