if you react 1.20g of copper metal with excess silver nitrate, how many moles of silver metal are produced?
Cu + 2AgNO3 ==> Cu(NO3)2 + 2Ag
mols Cu = grams/atomic mass = ?
mols Ag = 2x mols Cu
To find the number of moles of silver metal produced, you need to consider the chemical reaction between copper and silver nitrate. The balanced equation for the reaction is as follows:
Cu + 2AgNO3 → Cu(NO3)2 + 2Ag
From the balanced equation, you can see that one mole of copper (Cu) reacts with two moles of silver nitrate (AgNO3) to produce one mole of copper(II) nitrate (Cu(NO3)2) and two moles of silver (Ag).
Now, let's calculate the number of moles of copper (Cu) in 1.20 grams. You need to know the molar mass of copper (Cu) to do this calculation. The molar mass of copper is approximately 63.55 g/mol.
Number of moles of copper (Cu) = Mass of copper (Cu) / Molar mass of copper (Cu)
= 1.20 g / 63.55 g/mol
Next, since copper is reacting with an excess of silver nitrate, copper is the limiting reagent. Therefore, the number of moles of silver metal produced will be equal to the number of moles of copper.
Number of moles of silver metal produced = Number of moles of copper (Cu)
= 1.20 g / 63.55 g/mol
Performing the calculation, the number of moles of silver metal produced is approximately 0.0189 mol.
Therefore, when you react 1.20g of copper metal with excess silver nitrate, approximately 0.0189 moles of silver metal are produced.