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. :-)
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?)
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