Final answer:
Using stoichiometry, 5.00 g of silver nitrate would produce 4.21 g of silver chloride when reacting with an excess amount of barium chloride, but this value is not among the provided options, suggesting a possible typo in the question or answer choices.
Step-by-step explanation:
To determine how many grams of silver chloride are produced from 5.00 g of silver nitrate reacting with an excess amount of barium chloride, we use stoichiometry. The balanced chemical equation is 2AgNO₃ + BaCl₂ -> 2AgCl + Ba(NO₃)₂. Since the molar mass of AgNO3 is 169.88 g/mol, we first convert the mass of AgNO₃ to moles:
5.00 g AgNO₃ x (1 mol AgNO₃ / 169.88 g AgNO₃) = 0.0294 mol AgNO₃
The stoichiometry of the reaction shows that 2 moles of AgNO3 produce 2 moles of AgCl, which means the mole ratio is 1:1. Therefore, the moles of silver chloride produced will also be 0.0294 moles. Next, we convert moles of AgCl to grams using its molar mass (143.32 g/mol):
0.0294 mol AgCl x (143.32 g AgCl / 1 mol AgCl) = 4.21 g AgCl
However, since this value is not an option, there must have been a typo in the question statement or in the options provided. If any of the answer choices were correct based on the given information and calculations, option (c) 2.78g would be the closest, but it is still not correct.