Final answer:
From 55.8 g of silver nitrate, 45.6 g of lead II nitrate can be produced in the given reaction.
Step-by-step explanation:
In the given reaction, 1 mole of lead chloride (PbCl2) is produced from 2 moles of silver nitrate (AgNO3). To calculate the mass of lead chloride produced from 55.8 g of silver nitrate, we need to convert the mass of silver nitrate to moles and then use the stoichiometric ratio.
The molar mass of silver nitrate is 169.88 g/mol, so 55.8 g is equal to 0.328 moles of silver nitrate. According to the stoichiometry, 2 moles of silver nitrate produce 1 mole of lead chloride. Therefore, 0.328 moles of silver nitrate will produce 0.164 moles of lead chloride. To find the mass of lead chloride, we multiply the moles by its molar mass. The molar mass of lead chloride is 278.1 g/mol, so 0.164 moles is equal to 45.6 g of lead chloride.