For the chemical reaction, sodium hydroxide + copper (II) chloride = "robin's egg blue" precipitate, do the following: write the balanced molecular, balanced total ionic, and balanced net ionic equations; determine the Limiting Reactant (LR) for each chemical reaction, assuming that the concentration of each solution is 0.500M and the volume of each solution is 50.00mL; determine the theoretical yield of the product of interest; and, determine the actual yield of the product of interest if the percent yield was found to be 87.3%.
Chemical equation:
2NaOH(aq) + CuCl2(aq) --> 2NaCl(aq) + Cu(OH)2(s)
Chemical reaction: Double replacement
I posted this before, but the response said to tell what I know so I have done that now. I know it's not much. I'm kind of stuck on these types of problems. I don't know how to solve for the limiting reactant with concentrations and volume. I also don't understand theoretical yields and percent yields. Could you please explain how to solve this? Thank you so much!
What you say you don't understand is far more than can be explained here, it is almost two weeks of normal instruction. I suggest this:
a) http://www.chemteam.info/Equations/Net-Ionic-Equation.html
b) https://www.khanacademy.org/science/chemistry/chemical-reactions-stoichiome/limiting-reagent-stoichiometry/v/stoichiometry-limiting-reagent
c) http://www.sparknotes.com/chemistry/stoichiometry/realworldreactions/section2.rhtml
Those three links cover your not understanding, go through them, they should help.
You may need a tutor (quickly) before you are lost.
Sure! I will walk you through each step of solving this problem.
Step 1: Write the balanced molecular equation.
The given equation is already balanced:
2NaOH(aq) + CuCl2(aq) → 2NaCl(aq) + Cu(OH)2(s)
Step 2: Write the balanced total ionic equation.
In a total ionic equation, soluble ionic compounds are dissociated into their respective ions. The only insoluble compound is shown with its molecular formula.
2Na+(aq) + 2OH-(aq) + Cu2+(aq) + 2Cl-(aq) → 2Na+(aq) + 2Cl-(aq) + Cu(OH)2(s)
Step 3: Write the balanced net ionic equation.
In a net ionic equation, we eliminate the spectator ions, which are ions that appear on both sides of the equation cancel each other out. Generally, the spectator ions are monoatomic ions.
2OH-(aq) + Cu2+(aq) → Cu(OH)2(s)
Step 4: Determine the Limiting Reactant (LR) for each chemical reaction.
To determine the limiting reactant, we compare the stoichiometry of the reactants with their concentrations and volumes. In this case, the concentrations of each solution are 0.500M, and the volume of each solution is 50.00mL (hence, 0.0500L) for both reactants.
The number of moles of each reactant can be calculated using the formula:
n = c × V
Where:
n = number of moles
c = concentration
V = volume in liters
Let's calculate the number of moles for each reactant:
Moles of NaOH = 0.500M × 0.0500L = 0.0250 mol
Moles of CuCl2 = 0.500M × 0.0500L = 0.0250 mol
Since the stoichiometric coefficient of NaOH and CuCl2 is 1:1, we can see that the moles of each reactant are equal. Therefore, in this case, both NaOH and CuCl2 are the limiting reactants.
Step 5: Determine the theoretical yield of the product of interest.
The theoretical yield is the maximum amount of product that can be obtained from the limiting reactant, assuming complete reaction. From the balanced net ionic equation, we can see that 1 mole of Cu(OH)2 is produced per 1 mole of CuCl2.
Therefore, the theoretical yield of Cu(OH)2 is equal to the number of moles of the limiting reactant (CuCl2) in this case, which is 0.0250 moles.
Step 6: Determine the actual yield of the product of interest.
The actual yield is the amount of product actually obtained from the reaction. We are given that the percent yield is 87.3%. Percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100.
Let's solve for the actual yield:
Percent yield = (Actual yield / Theoretical yield) × 100
Rearranging the equation:
Actual yield = (Percent yield / 100) × Theoretical yield
Actual yield = (87.3 / 100) × 0.0250 mol ≈ 0.0218 mol
Therefore, the actual yield of Cu(OH)2 is approximately 0.0218 moles.
And that's it! You have now solved the given problem by writing the balanced equations, identifying the limiting reactant, calculating the theoretical yield, and determining the actual yield using the given percent yield. If you have any further questions, feel free to ask!