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Topic: Theoretical Yield Setup  (Read 5598 times)

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Offline aquablue8

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Theoretical Yield Setup
« on: February 27, 2011, 11:55:28 AM »
Hi,

Here is my problem and this is what I have attempted:


What is the theoretical yield of calcium carbonate if 2.07 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 hundreths place, and round your answer to the hundredths place.

CaCl2•2H2O(aq)+Na2CO3(aq)→CaCO3(aq)+2NaCl(aq)+2H2O(l)

CaCl2 is the reactant? So I took the molar mass of it, 110.98g/mol, and divided 2.07g CaCl2 by it to get .o2 g CaCl2??

Offline aquablue8

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Re: Theoretical Yield Setup
« Reply #1 on: February 27, 2011, 12:37:59 PM »
Here is what I have attempted: 2.07gCaCl2 x 1 mol Ca/40.08g Ca x 2 mol Cl/2 mol Ca x 70.9g Cl2/ 1 mol Cl2 = 3.62 g CaCl2??

Offline sjb

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Re: Theoretical Yield Setup
« Reply #2 on: February 27, 2011, 01:01:44 PM »
Hi,

Here is my problem and this is what I have attempted:


What is the theoretical yield of calcium carbonate if 2.07 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 hundreths place, and round your answer to the hundredths place.

CaCl2•2H2O(aq)+Na2CO3(aq)→CaCO3(aq)+2NaCl(aq)+2H2O(l)

CaCl2 is the reactant? So I took the molar mass of it, 110.98g/mol, and divided 2.07g CaCl2 by it to get .o2 g CaCl2??

Right sort of idea, but check your units, and the molecular weight of your reactant

Online Borek

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Re: Theoretical Yield Setup
« Reply #3 on: February 27, 2011, 04:02:34 PM »
Here is what I have attempted: 2.07gCaCl2 x 1 mol Ca/40.08g Ca x 2 mol Cl/2 mol Ca x 70.9g Cl2/ 1 mol Cl2 = 3.62 g CaCl2??

or in short version:

2.07gCaCl2 = 3.62 g CaCl2??

2=3, interesting result, but obviously incorrect.
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