Question

How many grams of silver metal would you expect to be produced when 250.0 grams of an 80.0% by mass silver nitrate solution reacts with excess copper metal?

Cu(s) + 2AgNO3(ag) -> 2Ag(s) + Cu(NO3)2(ag)

Please show how to solve too!

Answer 1

To determine the number of grams of silver metal produced in the reaction, we need to use stoichiometry and the given information.

Start by calculating the amount of silver nitrate (AgNO3) in the solution:

Mass of silver nitrate = 250.0 grams × 0.80 = 200.0 grams

Convert the mass of silver nitrate to moles using its molar mass:

Molar mass of AgNO3 = 107.87 grams/mol + 14.01 grams/mol + (3 × 16.00 grams/mol) = 169.87 grams/mol

Moles of AgNO3 = 200.0 grams / 169.87 grams/mol ≈ 1.176 moles

Based on the balanced equation, the stoichiometric ratio between silver nitrate and silver metal is 2:2. This means that for every 2 moles of AgNO3, we will produce 2 moles of Ag.

Calculate the moles of silver metal produced:

Moles of Ag = 1.176 moles × (2 moles Ag / 2 moles AgNO3) = 1.176 moles

Finally, convert the moles of silver to grams using the molar mass of silver:

Molar mass of Ag = 107.87 grams/mol

Grams of Ag = 1.176 moles × 107.87 grams/mol ≈ 126.4 grams

Therefore, you would expect approximately 126.4 grams of silver metal to be produced when 250.0 grams of an 80.0% by mass silver nitrate solution reacts with excess copper metal.

Step-by-step explanation: