Posted by Anonymous on Wednesday, October 3, 2012 at 12:39am.
CaCO3 + 2HCl ==> CaCl2 + H2O + CO2
Convert 27.0 g CaCO3 to mols.
Convert 14.0 g HCl to mols.
Use the coefficients to convert mols CaCO3 to mols CaCl2.
Do the same for HCl.
It is likely that the two values will not be the same (mols CaCl2 that is) and one of them must be wrong. The correct answer in limiting reagent is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Using the smaller value convert mols CaCl2 to grams. g = mols x molar mass.
To find the amount of excess reagent unreacted do this.
Using the coefficients convert mols of the limiting reagent to mols of the "other" reagent. That gives the amount of the non-limiting reagent USED. Convert to grams and subtract from the initial amount. The difference is the amount of excess reagent unreacted.
Post your work if you get stuck.
I got 0.270 mol CaCO3, 0.384 mol HCl, and 0.270 mol CaCl2. The LR is HCl. I don't know what to do next.
You're ok for part of it. 0.270 mol CaCO3 is right. 0.384 mol HCl is right. But you will not form 0.170 mol CaCl2. You have LR of HCl but you must have arrived at that answer incorrectly.
Convert mols CaCO3 to mols CaCl2 by
0.270 x (1 mol CaCl2/1 mol CaCO3) = 0.270.
Convert mols HCl to mols CaCl2 by
0.384 mols HCl x (1 mol CaCl2/2 mol HCl) = 0.192 mols CaCl2. Thus mols CaCl2 by HCl is the smaller value of 0.279 and 0.192 and 0.192 wins as the LR.
So you will form 0.385 x 1/2 = 0.192 mols CaCl2. Convert that to grams by g = mols x molar mass CaCl2.
To find how much CaCO3 is left (the excess reagent), convert mols HCl to mols CaCO3. That's 0.384 x (1 mol CaCO3/2 mol HCl) = 0.384 x 1/2 = 0.192 mols CaCO3 used. Grams CaCO3 used = 0.192 x molar mass CaCO3 = approximately 19 g.
You started with 27.0, you've used about 19 so you must have 27-19 = about 8 g CaCO3 left over.
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