Final answer:
The mass of lead (II) chloride produced is 18.7 grams when 14 grams of barium chloride reacts with lead (II) nitrate.
Step-by-step explanation:
In this reaction, barium chloride (BaCl2) reacts with lead (II) nitrate (Pb(NO3)2) to form lead (II) chloride (PbCl2). The balanced chemical equation for this reaction is:
BaCl2 + Pb(NO3)2 → PbCl2 + Ba(NO3)2
To find the mass of lead (II) chloride produced, we need to calculate the molar mass of lead (II) chloride and use stoichiometry. The molar mass of PbCl2 is 278.1 g/mol, and the molar mass of BaCl2 is 208.2 g/mol.
- Calculate the moles of BaCl2: 14 g / 208.2 g/mol = 0.0672 moles
- Since the molar ratio between BaCl2 and PbCl2 is 1:1, the moles of PbCl2 produced will also be 0.0672 moles.
- Calculate the mass of PbCl2: 0.0672 moles × 278.1 g/mol = 18.7 g