Question

How many grams of each species will be present at the end when 0.500 g of copper is added to 2.500 g of AgNO₃

Cu(s) + 2AgNO₃ (aq)→2Ag(s) + Cu (NO₃)₂ (aq)

Answer 1

The question involves stoichiometry and after determining the limiting reactant, which is AgNO3, we find that 1.588 grams of silver (Ag) and 1.382 grams of copper (II) nitrate (Cu(NO3)2) are formed at the end of the reaction.

Step-by-step explanation:

The question is asking to determine the amounts of each substance present after the reaction between solid copper and silver nitrate is complete. The balanced chemical equation for this reaction is:

Cu (s) + 2AgNO3 (aq) → Cu(NO3)2 (aq) + 2Ag (s)

According to the stoichiometry of the reaction, one mole of copper reacts with two moles of silver nitrate to produce one mole of copper (II) nitrate and two moles of silver. To solve this problem, we need to follow these steps:

1. Determine the moles of copper used. The molar mass of copper (Cu) is 63.55 g/mol.
2. Calculate the moles of silver nitrate (AgNO3), given that its molar mass is 169.87 g/mol.
3. Use the mole ratio from the balanced equation to find out which reactant is the limiting reactant.
4. Use the limiting reactant to determine the amounts of products using the balanced equation.

To find the moles of copper, we divide the given mass by the molar mass:

0.500 g Cu * (1 mol Cu / 63.55 g Cu) = 0.00787 mol Cu

To find the moles of silver nitrate:

2.500 g AgNO3 * (1 mol AgNO3 / 169.87 g AgNO3) = 0.01472 mol AgNO3

The mole ratio between Cu and AgNO3 is 1:2, so we would need twice as many moles of AgNO3 to react completely with the moles of Cu we have. However, we have less than the required amount of AgNO3, making AgNO3 the limiting reactant.

Since AgNO3 is limiting, all of it will be consumed to form products. Using the mole ratio from the balanced equation, we can calculate the grams of products formed:

0.01472 mol AgNO3 * (2 mol Ag / 2 mol AgNO3) * (107.87 g Ag / 1 mol Ag) = 1.588 g Ag (formed)

0.01472 mol AgNO3 * (1 mol Cu(NO3)2 / 2 mol AgNO3) * (187.55 g Cu(NO3)2 / 1 mol Cu(NO3)2) = 1.382 g Cu(NO3)2 (formed)

Thus, at the conclusion of the reaction, there will be approximately 1.588 g of silver and 1.382 g of copper (II) nitrate. The copper used will have been completely consumed.