Stoichiometry;
Find the percent yield of solid calcium carbonate made when 5.0 mL of 1.0 M Sodium Carbonate reacts with 0.40g Calcium Chloride
Balanced Equation; Na2(CO3)+CaCl2-->2NaCl+CaCo3
could you please help me step by step to find the theoretical yield as well as the limiting reactant/reagent(once the limiting reactant is found what is done with it?)
To find the percent yield, theoretical yield, and the limiting reactant in this reaction, follow these steps:
Step 1: Write down the balanced equation
The balanced equation is already given as Na2(CO3) + CaCl2 -> 2NaCl + CaCO3.
Step 2: Calculate the number of moles of each reactant
Given:
- Volume of Sodium Carbonate solution: 5.0 mL = 0.005 L
- Concentration of Sodium Carbonate solution: 1.0 M
- Mass of Calcium Chloride: 0.40 g
First, calculate the number of moles of Sodium Carbonate:
moles of Sodium Carbonate = concentration * volume
moles of Sodium Carbonate = 1.0 M * 0.005 L
Next, calculate the number of moles of Calcium Chloride:
moles of Calcium Chloride = mass / molar mass
First, find the molar mass of Calcium Chloride:
CaCl2 = 40.08 g/mol (molar mass of Ca) + 2 * 35.45 g/mol (molar mass of Cl)
moles of Calcium Chloride = 0.40 g / (40.08 g/mol + 2 * 35.45 g/mol)
Step 3: Determine the limiting reactant
To determine the limiting reactant, compare the moles of Sodium Carbonate to Calcium Chloride. The reactant that produces fewer moles of CaCO3 is the limiting reactant.
Convert the moles of Sodium Carbonate to moles of CaCO3 using the stoichiometry of the balanced equation. From the balanced equation, it is known that 1 mole of CaCO3 is produced for every 1 mole of Sodium Carbonate.
moles of CaCO3 produced from Sodium Carbonate = moles of Sodium Carbonate
Convert the moles of Calcium Chloride to moles of CaCO3 using the stoichiometry of the balanced equation. From the balanced equation, it is known that 1 mole of CaCO3 is produced for every 1 mole of Calcium Chloride.
moles of CaCO3 produced from Calcium Chloride = moles of Calcium Chloride
Compare the moles of CaCO3 produced by each reactant. The reactant that produces fewer moles of CaCO3 is the limiting reactant.
Step 4: Calculate the theoretical yield
The theoretical yield is the maximum amount of CaCO3 that can be produced based on the limiting reactant.
From the previous step, you have determined the moles of the limiting reactant. Now using the stoichiometry from the balanced equation, you can calculate the moles of CaCO3 produced.
moles of CaCO3 produced = moles of limiting reactant
Finally, calculate the mass of CaCO3 using the moles and the molar mass of CaCO3.
mass of CaCO3 = moles of CaCO3 * molar mass of CaCO3
Step 5: Calculate the percent yield
The percent yield can be calculated by dividing the actual yield (mass obtained in the experiment) by the theoretical yield and multiplying by 100.
percent yield = (actual yield / theoretical yield) * 100
Note: The actual yield is not provided in the problem statement. It is typically determined through experimentation.
Once the limiting reactant is identified, you can use it to calculate the theoretical yield. The theoretical yield provides a benchmark to compare the actual yield against and determine the percent yield.
Sure! To find the percent yield of solid calcium carbonate, you need to follow a few steps. Let's break it down step by step:
Step 1: Write the balanced equation
The balanced equation you provided is correct:
Na2(CO3) + CaCl2 → 2NaCl + CaCO3
Step 2: Calculate the molar amount of each reactant
To determine the limiting reactant and the theoretical yield, we need to compare the molar amounts of each reactant. First, let's calculate the number of moles for each reactant.
Given:
- Volume of sodium carbonate (Na2(CO3)) = 5.0 mL
- Concentration of sodium carbonate = 1.0 M
- Mass of calcium chloride (CaCl2) = 0.40 g
- Molar mass of Na2(CO3) = 105.99 g/mol
- Molar mass of CaCl2 = 110.98 g/mol
1. Calculate moles of Na2(CO3):
Moles of Na2(CO3) = (Volume in liters) x (Concentration in mol/L)
= (5.0 mL / 1000 mL/L) x (1.0 mol/L)
= 0.005 mol
2. Calculate moles of CaCl2:
Moles of CaCl2 = Mass / Molar mass
= 0.40 g / 110.98 g/mol
= 0.0036 mol (rounded to four decimal places)
Step 3: Determine the limiting reactant
To determine the limiting reactant, compare the mole ratios in the balanced equation. The reactant that produces the lesser moles of product will be the limiting reactant.
From the balanced equation: 1 mol Na2(CO3) produces 1 mol CaCO3
1 mol CaCl2 produces 1 mol CaCO3
In this case, the moles of Na2(CO3) = 0.005 mol and the moles of CaCl2 = 0.0036 mol. The limiting reactant is the one that produces fewer moles of CaCO3. Therefore, CaCl2 is the limiting reactant.
Step 4: Determine the theoretical yield
Now that we know CaCl2 is the limiting reactant, we can determine the theoretical yield of CaCO3. We need to calculate the moles of CaCO3 that can be produced using the limiting reactant.
From the balanced equation: 1 mol CaCl2 produces 1 mol CaCO3
Since the moles of CaCl2 = 0.0036 mol, the moles of CaCO3 produced will also be 0.0036 mol.
The molar mass of CaCO3 is 100.09 g/mol. Multiply the moles of CaCO3 by the molar mass to get the theoretical yield in grams:
Theoretical yield = Moles of CaCO3 x Molar mass of CaCO3
= 0.0036 mol x 100.09 g/mol
= 0.36 g
So, the theoretical yield of solid calcium carbonate is 0.36 grams.
Step 5: Calculate the percent yield
The percent yield is calculated by dividing the actual yield (which is not given) by the theoretical yield found in the previous step, and multiplying by 100.
Percent yield = (Actual yield / Theoretical yield) x 100
Since the actual yield is not provided in your question, it's not possible to determine the percent yield without that information.
That's it! By following these steps, you have determined the theoretical yield of solid calcium carbonate and found the limiting reactant/reagent.
The limiting reagent problems are really just two ordinary stoichiometry problems solved more or less together. Here is how it is done. The first step is to write and balance the equation which you have supplied.
CaCl2 + Na2CO3 ==> CaCO3 + 2NaCl
Step2a. Convert g CaCl2 to mols. mols = grams/molar mass = approx 0.0036 but you need to go through ALL these calculations and do them more accurately.
2b. Convert M and L Na2CO3 to mols. mols = M x L = approx 0.005
Step 3a. Now using the coefficients in the balanced equation, convert each of the mols in 2a and 2b to mols of the product, CaCO3. For CaCl2 that's
0.0036 mols CaCl2 x (1 mol CaCO3/1 mol CaCl2) = 0.0036 x (1/1) = 0.0036 mol CaCO3 approx.
3b. Do the same for Na2CO3.
0.005 mol Na2CO3 x (1 mol CaCO3/1 mol Na2CO3) = 0.005 x (1/1) = approx 0.005 mols CaCO3.
3c. You see the values for mols product formed don't agree so one of them must be wrong; the correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Step 4. The above takes care of two things. One is that you now know mols of the product formed (in this case 0.0036 mols CaCO3) and the other is that the limiting reagent has been identified. To determine how much CaCO3 is produced in grams it is simply
grams = mols x molar mass = 0.0036 x 100g CaCO3/mol CaCO3 = 0.36 g CaCO3 produced. This is the theoretical yield and the end of the problem. This particular problem doesn't ask you to do anything with the limiting reagent. So you say, "Why did I identify it?" and the answer is mostly because the problem asked for it to be identified. Some teachers identify the limiting reagent FIRST and use that to determine the stoichiometry of the problem. I don't do it that way. I work out the theoretical yield and in doing so that identifies the limiting reagent. Hope this helps. Print this out and save it. This procedure will work 99.9% of the limiting reagent problems you are likely to encounter. Sometimes a problem will ask how much of the "other non-limiting reagent" remains unreacted and that's just another ordinary stoichiometry problem. If you were asked that in this problem you would do this. The "other" reagent is Na2CO3. How much was used.
0.0036 mols CaCl2 x (1 mol Na2CO3/1 mol CaCl2) = 0.0036 x 1/1 = 0.0036 mols Na2CO3 used. You had 0.005 initially so you must have 0.005-0.0036 = 0.0014 and that converted mL left is M = mols/L.
1M = 0.0014/L and L = 0.0014/1M = 0.0014L and that converted to mL = 1.4mL. So you used up 3.6 of the 5.00 mL to be left with 1.4 mL of the 1M Na2CO3 unreacted. :-)