Answer:
For 0.353 moles AgNO3, we'll have 0.353 moles AgCl
Explanation:
How many moles of AgCl will be produced from 60.0g AgNO3 assuming NaCl is available in excess.
Step 1: Data given
Mass of AgNO3 = 60.0 grams
Molar mass AgNO3 = 169.87 g/mol
NaCl is in excess, so AgNO3 is the limiting reactant
Step 2: The balanced equation
AgNO3 + NaCl → AgCl + NaNO3
Step 3: Calculate moles AgNO3
Moles AgNO3 = mass AgNO3 / molar mass AgNO3
Moles AgNO3 = 60.0 grams / 169.87 g/mol
Moles AgNO3 = 0.353 moles
Step 4: Calculate moles AgCl
For 1 mol AgNO3 we need 1 mol NaCl to produce 1 mol AgCl and 1 mol NaNO3
For 0.353 moles AgNO3, we'll have 0.353 moles AgCl
