How many grams of silver nitrate will be needed to produce 8.6 g of silver?

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) =