If 50.0 grams of copper metal is added to a silver nitrate solution, what mass of silver is produced if copper 2 nitrate is also formed?
To solve this problem, we need to understand the concept of a chemical reaction and the concept of stoichiometry.
When copper metal is added to a silver nitrate solution, a chemical reaction takes place. Copper reacts with silver nitrate to form copper(II) nitrate and silver, according to the following balanced chemical equation:
2 AgNO3 + Cu → Cu(NO3)2 + 2 Ag
From the balanced equation, we can see that 2 moles of silver (Ag) are produced for every 1 mole of copper (Cu). To find the mass of silver produced, we'll follow these steps:
Step 1: Determine the number of moles of copper (Cu).
Given that the mass of copper is 50.0 grams and the molar mass of copper is 63.55 g/mol, we can use the formula:
moles = mass / molar mass
moles of Cu = 50.0 g / 63.55 g/mol = 0.786 mol
Step 2: Use the mole ratio from the balanced equation to find the moles of silver (Ag) produced.
From the balanced equation, we know that the mole ratio of copper to silver is 1:2. Therefore, the moles of silver produced can be calculated as:
moles of Ag = 2 * moles of Cu = 2 * 0.786 mol = 1.572 mol
Step 3: Calculate the mass of silver (Ag) produced.
To find the mass of silver produced, we'll use the formula:
mass = moles * molar mass
The molar mass of silver (Ag) is 107.87 g/mol, so we have:
mass of Ag = 1.572 mol * 107.87 g/mol = 169.5 g
Therefore, if 50.0 grams of copper metal is added to a silver nitrate solution, approximately 169.5 grams of silver is produced when copper(II) nitrate is also formed.