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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?

Sagot :

zcathy

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

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