How many grams of solid copper must react with silver nitrate, AgNO3(aq), to produce 5.5 g of solid silver and copper(II) nitrate, Cu(NO3)s(aq)?
Same stoichiometry problem.
5.2
To determine the number of grams of solid copper needed to react with silver nitrate, we can use the concept of stoichiometry. This involves balancing the chemical equation and using the molar ratios of the reactants and products to find the answer.
The balanced chemical equation for the reaction between copper and silver nitrate is:
Cu(s) + 2AgNO3(aq) → 2Ag(s) + Cu(NO3)2(aq)
From the balanced equation, we can see that 1 mole of copper reacts with 2 moles of silver nitrate to produce 2 moles of silver.
To find the molar mass of silver, we can look it up on the periodic table. The molar mass of silver (Ag) is approximately 107.87 g/mol.
Therefore, 2 moles of silver would be equal to (2 mol) x (107.87 g/mol) = 215.74 g.
Since the reaction produces 5.5 g of silver, we can set up the following equation:
(215.74 g Ag) / (2 mol Ag) = (5.5 g Ag) / (x mol Cu)
Solving for x, we can find the number of moles of copper needed.
x mol Cu = (5.5 g Ag) × (2 mol Ag) / (215.74 g Ag)
x mol Cu ≈ 0.101 mol Cu
To convert the moles of copper to grams, we can use the molar mass of copper (Cu) from the periodic table, which is approximately 63.55 g/mol.
Grams of copper = (0.101 mol Cu) × (63.55 g/mol Cu)
Grams of copper ≈ 6.45 g
Therefore, approximately 6.45 grams of solid copper must react with silver nitrate to produce 5.5 grams of solid silver and copper(II) nitrate.