Final answer:
To produce 80.0 grams of silver chloride from the reaction of silver nitrate with gold (III) chloride, you would need 0.558 moles of silver nitrate according to the stoichiometry of the balanced chemical equation.
Step-by-step explanation:
To determine how many moles of silver nitrate are needed to react with gold (III) chloride to produce 80.0 grams of silver chloride, stoichiometry is used.
First, the molar mass of silver chloride, AgCl, is calculated, which is 143.32 g/mol.
Then, we find the number of moles of AgCl that 80.0 grams correspond to by dividing the mass by the molar mass: 80.0 g ÷ 143.32 g/mol = 0.558 moles of AgCl.
According to the balanced chemical equation, 3 moles of AgNO3 react to produce 3 moles of AgCl.
Therefore, the mole ratio of AgNO3 to AgCl is 1:1. As such, to produce 0.558 moles of AgCl, you would need 0.558 moles of AgNO3.