Final answer:
To produce 9.272 moles of AgCl, one would need 541.64 grams of NaCl based on a one-to-one stoichiometric ratio with AgNO3 and the molar mass of NaCl.
Step-by-step explanation:
To calculate how many grams of NaCl are needed to make 9.272 moles of AgCl, we look at the stoichiometry of the reaction where silver nitrate (AgNO3) reacts with sodium chloride (NaCl) to form silver chloride (AgCl) and sodium nitrate (NaNO3). The reaction is stoichiometrically one-to-one for all reactants and products. To find the required amount of NaCl, we multiply the number of moles by the molar mass of NaCl.
Calculation steps
Confirm the stoichiometry of the reaction: NaCl + AgNO3 → AgCl + NaNO3 (1:1:1:1 ratio).
Calculate the molar mass of NaCl (22.99 g/mol for Na + 35.45 g/mol for Cl = 58.44 g/mol for NaCl).
Multiply the number of moles of AgCl by the molar mass of NaCl to get the mass of NaCl needed.
Mass of NaCl = moles of AgCl × molar mass of NaCl = 9.272 moles × 58.44 g/mol = 541.64 grams of NaCl.
Therefore, to make 9.272 moles of AgCl, we need 541.64 grams of NaCl.