Question

you have just purchased a storage shed that has 10,000kg of pure silver nitrate solution. you have an idea: "can I run a single replacement reaction to obtain pure silver and sell the silver?"

you know the price of silver on the metal exchange is about $594.79/kg. then, you know that pure aluminum will react with solver nitrate to produce pure silver. aluminum nitrate will be the other product. you must first purchase enough aluminum foil from the store to run the reaction. aluminum foil sells for about $.30/kg

question: using stiochiometry, calculate the amount of silver that is produced; note: make sure you conver kg to g prior to working problem

Answer 1

Given

10000 Kg of silver nitrate = 10⁷ g of Silver nitrate

Molar mass of silver nitrate = 170 g/mol

No. of moles of Silver nitrate = mass / Molar mass = 10⁷ g / 170 g/mol = 58823.53 mole

Al + AgNO₃ = Al(NO₃) + Ag

so 1 mole of AgNO₃wil give 1 mole of Ag and will require 1 mole of Al(NO₃)

so 58823.53 moles of silver will be produced

Molar mass of silver = 108 g/mol

Mass of silver produced = No. of moles * Molar mass of Silver= 588223.53 moles * 108 g/mol = 6352941.2 g of Ag

Amount of silver produced = 6353 Kg