Answer:
124 g (3 sig figs)
or
124.011 g (6 sig figs
Explanation:
Step 1: Calculate g/mol for AgNO₃
Ag - 107.868 g/mol
N - 14.01 g/mol
O - 16.00 g/mol
107.868 + 14.01 + 16.00(3) = 169.878 g/mol
Step 2: Multiply 0.73 moles by molar mass
0.73 mol (169.979 g/mol)
124 grams of AgNO₃
