Hi, I have a few questions that I am struggling with. Any help would be greatly appreciated.
Question 1
Calculate the molar mass of calcium chloride dihydrate, CaCl2 • 2H2O.
Please use the Periodic Table values rounded to the hundredths place for your calculation and round your answer to the hundredths place and remember, the "dot" (•) means closely associated, reprsenting the ADDITION of two waters of hydration to the calcium chloride formula unit.
Question 2
What is the theoretical yield of calcium carbonate if 2.97 grams of calcium chloride dihydrate reacts with excess sodium carbonate according to the balanced chemical reaction shown below?
Please use molar mass values calculated and rounded to the hundredths place, and round your answer to the hundredths place.
Question 3
What is the percent yield of calcium carbonate if your theoretical yield was 2.07 grams, and your actual yield was 1.46 grams, from the balanced chemical reaction shown below?
Please round your answer to the tenths place.
CaCl2•2H20(aq)+Na2CO3(aq)->CaCO3(aq)+2NaCL(aq)+2H20(I)
We don't do your homework for you but we'll be glad to help your through it. First, what do you not understand about calculating the molar mass of CaCl2.2H2O? That's pretty fundamental.
Also, it's better if you post one question per post. That way several tutors may tackle a problem whereas one person may not have the time to answer all three.
#1. See response above.
#2. You have the equation. Convert 2.97 g CaCl2.2H2O to moles. moles = grams/molar mass. Next, using the coefficients in the balanced equation, convert moles CaCl2.2H2O to moles CaCO3. Then convert moles CaCO3 to grams. grams = moles x molar mass.
#3.
%yield = [actual yield/theoretical yield]*100 = ??
Post your work if you need further assistance.
I apologize about the length of the message. This will be the last long one I post.
To figure out the molar mass of CaCl2•H2O (rounding to the hundredths place), I did
Ca = 40.08
Cl2 = (35.45) x 2 = 70.9
2H2O = (18.02) x 2 = 36.04
40.08 + 70.90 + 36.04 = 147.02 g/mol
2.97g CaCl•2H2O x (1 mol CaCl•2H2O/147.02g CaCl•2H2O) x (1 mol CaCO3/1 mol CaCl•2H2O) x (100.09/1 mol CaCO3) = 2.04 g
(The coefficients were all one, so I used 1 mol in all the neccessary conversion factors.)
Now, the third question asked "What is the percent yield of calcium carbonate if your theoretical yield was 2.07 grams" even though I came out with 2.04 g as my theoretical yield.
So, percentage yield, (1.46/2.07) X 100 = 70.5%.
Was I accurate in all these equations? If not, could you guide me to where I need to make a modification? Thanks!
Question 1:
To calculate the molar mass of calcium chloride dihydrate (CaCl2 • 2H2O), we need to find the molar mass of each individual element and then add them together.
The molar mass of calcium (Ca) is 40.08 g/mol (from the periodic table).
The molar mass of chlorine (Cl) is 35.45 g/mol (from the periodic table).
The molar mass of hydrogen (H) is 1.01 g/mol (from the periodic table).
The molar mass of oxygen (O) is 16.00 g/mol (from the periodic table).
Now, calculate the molar mass of calcium chloride:
2 x molar mass of chloride (Cl) = 2 x 35.45 g/mol = 70.90 g/mol
1 x molar mass of calcium (Ca) = 1 x 40.08 g/mol = 40.08 g/mol
Next, calculate the molar mass of water:
2 x (molar mass of hydrogen (H) + molar mass of oxygen (O)) = 2 x (1.01 g/mol + 16.00 g/mol) = 34.02 g/mol
Finally, add the molar masses of calcium chloride and water:
70.90 g/mol + 40.08 g/mol + 34.02 g/mol = 144.00 g/mol
Therefore, the molar mass of calcium chloride dihydrate, CaCl2 • 2H2O, is 144.00 g/mol.
Question 2:
To find the theoretical yield of calcium carbonate, we need to use the balanced chemical reaction and convert the given mass of calcium chloride dihydrate (CaCl2 • 2H2O) to moles, then use the stoichiometric coefficients to determine the moles of calcium carbonate, and finally convert back to grams using the molar mass of calcium carbonate.
The balanced chemical reaction is:
CaCl2•2H20(aq) + Na2CO3(aq) -> CaCO3(aq) + 2NaCL(aq) + 2H20(l)
Step 1: Convert the mass of calcium chloride dihydrate to moles.
Given mass of calcium chloride dihydrate = 2.97 g
Molar mass of calcium chloride dihydrate = 144.00 g/mol (from Question 1)
Number of moles = given mass / molar mass
Number of moles of calcium chloride dihydrate = 2.97 g / 144.00 g/mol = 0.02063 mol
Step 2: Use the stoichiometric coefficients to determine moles of calcium carbonate.
From the balanced chemical reaction, the stoichiometric coefficient of calcium carbonate (CaCO3) is 1.
Therefore, the number of moles of calcium carbonate produced is also 0.02063 mol.
Step 3: Convert moles of calcium carbonate to grams.
Molar mass of calcium carbonate = 100.09 g/mol (from the periodic table)
Mass of calcium carbonate = number of moles x molar mass
Mass of calcium carbonate = 0.02063 mol x 100.09 g/mol = 2.065 g
Therefore, the theoretical yield of calcium carbonate is 2.065 grams.
Question 3:
To find the percent yield of calcium carbonate, we need to use the given theoretical yield and actual yield.
Given:
Theoretical yield = 2.07 grams
Actual yield = 1.46 grams
Percent yield = (actual yield / theoretical yield) x 100
Percent yield = (1.46 g / 2.07 g) x 100 = 70.53%
Therefore, the percent yield of calcium carbonate is 70.5%.