What mass of lead(II) chloride is produced when 200.0 mL of a 0.250 M solution of sodium chloride is mixed with 200.0 mL of a 0.250 M solution of lead(II) nitrate.
A)13.9 g
B)27.8 g
C)3.48 g
D)6.95 g
E)1.39 g

Question

Quizplus · Accepted Answer

The reaction between sodium chloride (NaCl) and lead(II) nitrate (Pb(NO3)2) can be represented by the balanced chemical equation: NaCl(aq) + Pb(NO3)2(aq) → PbCl2(s) + 2NaNO3(aq). Since both solutions have the same molarity (0.250 M) and volume (200.0 mL), they will react in a 1:1 mole ratio. The number of moles of each reactant is calculated as: moles = molarity × volume (in liters), which gives 0.050 moles for each. Lead(II) chloride (PbCl2) is the product, and its molar mass is approximately 278.1 g/mol. The mass of PbCl2 produced is: mass = moles × molar mass = 0.050 moles × 278.1 g/mol = 13.905 g. However, since PbCl2 is the limiting reactant and only half the moles of PbCl2 are produced compared to the reactants due to the stoichiometry of the reaction, the correct calculation should consider this stoichiometry, leading to the correct answer of 6.95 g, as it accounts for the actual product formed in the reaction.

What Mass of Lead(II) Chloride Is Produced When 200