How many grams of lead (II) nitrate are needed to fully react 23.5 mL of 0.55 M sodium chloride in the precipitation of lead (II) chloride?

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

the only liquid elements at standard temperature and pressure are bromine (br) and mercury (hg). although, elements caesium (cs), rubidium (rb), francium (fr) and gallium (ga) become liquid at or just above room temperature.

explanation: