Answer:
13.5g of AgNO3 will be needed
Explanation:
Silver nitrate, AgNO3 contains 1 mole of silver, Ag, per mole of nitrate. To solve this problem we need to convert the mass of Ag to moles. Thee moles = Moles of AgNO3 we need. With the molar mass of AgNO3 we can find the needed mass:
Moles Ag-Molar mass: 107.8682g/mol-
8.6g * (1mol / 107.8682g) = 0.0797 moles Ag = Moles AgNO3
Mass AgNO3 -Molar mass: 169.87g/mol-
0.0797 moles Ag * (169.87g/mol) =
13.5g of AgNO3 will be needed
