Final answer:
To produce 84 g of silver chloride, 31.61 g of silver is needed, according to the stoichiometry of the balanced chemical equation Ag + Cl2 -> 2 AgCl.
Step-by-step explanation:
To calculate the mass of silver needed to react with chlorine to produce 84 g of silver chloride, we need to consider the stoichiometry of the balanced chemical equation for the reaction:
Ag + Cl2 → 2 AgCl
First, we calculate the molar mass of silver chloride (AgCl). Silver (Ag) has a molar mass of 107.87 g/mol, and chlorine (Cl) has a molar mass of 35.45 g/mol. Therefore, the molar mass of AgCl is 143.32 g/mol (107.87 g/mol + 35.45 g/mol).
To find the number of moles of AgCl that correspond to 84 g, we use the following calculation:
moles of AgCl = mass of AgCl / molar mass of AgCl
moles of AgCl = 84 g / 143.32 g/mol = 0.586 moles
The balanced equation shows that 2 moles of AgCl are produced for every 1 mole of Ag reacting. Therefore, to find the moles of Ag needed, we divide the moles of AgCl by 2:
moles of Ag = 0.586 moles AgCl x (1 mole Ag / 2 moles AgCl) = 0.293 moles Ag
Finally, to calculate the mass of Ag required:
mass of Ag = moles of Ag x molar mass of Ag
mass of Ag = 0.293 moles x 107.87 g/mol = 31.61 g of Ag
Therefore, the theoretical yield of silver needed is 31.61 g.