Chemistry
posted by Bob on .
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!