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What mass of lead(II) chloride forms in the reaction of 24.3 mL of 1.34 M lead(II) nitrate and 38.1 mL of 1.22 M sodium chloride?

Answer :

Final Answer:

The mass of lead(II) chloride formed in the reaction is X grams.

Explanation:

To determine the mass of lead(II) chloride formed in this chemical reaction, we can use the concept of stoichiometry and the given concentrations and volumes of reactants. First, we need to write the balanced chemical equation for the reaction:

Pb(NO3)2 + 2NaCl → PbCl2 + 2NaNO3

From the balanced equation, we can see that one mole of lead(II) nitrate (Pb(NO3)2) reacts with two moles of sodium chloride (NaCl) to produce one mole of lead(II) chloride (PbCl2).

Next, we calculate the number of moles of lead(II) nitrate and sodium chloride in the given volumes:

For lead(II) nitrate:

Moles = (Volume in L) x (Concentration in M)

Moles of Pb(NO3)2 = 0.0243 L x 1.34 M = 0.032382 moles

For sodium chloride:

Moles of NaCl = 0.0381 L x 1.22 M = 0.046482 moles

Since lead(II) nitrate reacts with sodium chloride in a 1:2 mole ratio, we need twice as many moles of sodium chloride as moles of lead(II) nitrate for complete reaction. However, the moles of sodium chloride exceed this requirement. Therefore, the limiting reactant is lead(II) nitrate, and it determines the amount of lead(II) chloride formed.

Now, we can calculate the moles of lead(II) chloride formed:

Moles of PbCl2 = Moles of Pb(NO3)2 = 0.032382 moles

Finally, we calculate the mass of lead(II) chloride using its molar mass (PbCl2 = 207.2 g/mol):

Mass = Moles x Molar Mass = 0.032382 moles x 207.2 g/mol ≈ 0.318 grams.

So, the mass of lead(II) chloride formed is approximately 0.318 grams.

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