when 10.0 ml of a .20 M Cu (NO3)2 solution is mixed with 15.0 mL of 0.20M NaOH, what is the theoretical yield (grams) of copper (II) hydroxide?
Write and balance the equation.
Convert Cu(NO3)2 to mols. mol = M x L = ?
Using the coefficients in the balanced equation convert mols Cu(NO3)2 to mols Cu(OH)2.
Now convert mols Cu(OH)2 to grams. g = mols x molar mass. This is the theoretical yield.
what's the balanced equation?
Cu(NO3)2 + 2NaOH ==> Cu(OH)2 + 2HNO3
To determine the theoretical yield of copper (II) hydroxide in this reaction, you need to calculate the amount of moles of Cu(NO3)2 and NaOH used and determine the stoichiometry of the reaction to find the limiting reactant.
Step 1: Calculate the moles of Cu(NO3)2:
Moles of Cu(NO3)2 = Volume (in L) × Concentration (in mol/L)
= 10.0 mL × (1 L / 1000 mL) × 0.20 mol/L
= 0.002 mol
Step 2: Calculate the moles of NaOH:
Moles of NaOH = Volume (in L) × Concentration (in mol/L)
= 15.0 mL × (1 L / 1000 mL) × 0.20 mol/L
= 0.003 mol
Step 3: Determine the stoichiometry of the reaction:
From the balanced chemical equation:
Cu(NO3)2 + 2NaOH → Cu(OH)2 + 2NaNO3
The stoichiometric ratio of Cu(NO3)2 to Cu(OH)2 is 1:1. This means that for every 1 mole of Cu(NO3)2, 1 mole of Cu(OH)2 is formed.
Step 4: Identify the limiting reactant:
To determine the limiting reactant, compare the moles of Cu(NO3)2 and NaOH. The reactant that produces fewer moles of Cu(OH)2 will be the limiting reactant.
Since 1 mole of Cu(NO3)2 produces 1 mole of Cu(OH)2, both reactants produce the same moles of Cu(OH)2. Therefore, there is no limiting reactant in this case.
Step 5: Calculate the theoretical yield of Cu(OH)2:
The theoretical yield is the amount of Cu(OH)2 that can be formed based on the stoichiometry of the balanced equation.
Moles of Cu(OH)2 = Moles of limiting reactant (Cu(NO3)2) = 0.002 mol
Molar mass of Cu(OH)2 = Atomic mass of Cu + 2 × atomic mass of H + 2 × atomic mass of O
= (63.55 g/mol) + 2 × (1.01 g/mol) + 2 × (16.00 g/mol)
= 97.55 g/mol
Theoretical yield of Cu(OH)2 = Moles of Cu(OH)2 × Molar mass of Cu(OH)2
= 0.002 mol × 97.55 g/mol
= 0.1951 g
Therefore, the theoretical yield of copper (II) hydroxide in this reaction is 0.1951 grams.