How Did My Professor Calculate 45 mmoles of Limiting Reagent in a Chemistry Lab?

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SUMMARY

The discussion centers on calculating the amount of limiting reagent in a chemistry lab experiment involving 50 mL of 1.02 M H3PO4 and 45 mL of 1 M Na2CO3. The professor determined that there are 45 mmoles of the limiting reagent, which the student initially miscalculated as 467 mmoles of Na2CO3. The key takeaway is that the mass of the solution does not equal the mass of the reagent dissolved, clarifying the confusion regarding the limiting reagent calculation.

PREREQUISITES
  • Understanding of molarity and its calculation
  • Knowledge of density and its application in mass calculations
  • Familiarity with stoichiometry in chemical reactions
  • Basic principles of limiting reagents in chemistry
NEXT STEPS
  • Review molarity calculations and conversions between moles and millimoles
  • Study the concept of limiting reagents and how to identify them in chemical reactions
  • Learn about density and its role in converting volume to mass
  • Practice stoichiometric calculations using different chemical equations
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Chemistry students, educators, and anyone involved in laboratory experiments requiring stoichiometric calculations and understanding of limiting reagents.

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Homework Statement



Hi. I'm reaady to panic right now so if anyone could please take the time to answer my question I would HUGELY appreciate it! I seriously need to get this done soon as I also have a test tomorrow and i just want to cry because I'm getting nowhere...

We did a lab experiment which involved 50 mL of 1.02 Molar H3PO4. The density was given to be 1.05 g/mL.

We also had 45 mL of 1 Molar Na2CO3 with a density of 1.10 g/mL.

We were asked to find the mass of both solutions. I got 52.5 g of H3PO4 and 49.5 g of Na2CO3.


Our professor gave us the value that there is 45 mmoles of limiting reagent. And I would assume the Na2CO3 is the limiting reagent because there is less of it.

But HOW IN THE WORLD did he get 45 mmoles?

This is how I'm trying to do it:

(49.5 g Na2CO3)/ (106 g Na2CO3)= .467 moles, or 467 mmoles of Na2CO3, and since there are the same number of moles of CO3 2- as Na2CO3, the mmoles of CO3 2- should also be 467 mmoles.

why am i SOOOOO far off?

PLEASE HELP! i have a million questions to do on this lab and i can't even understand why my numbers are the way they are so i can't even start TRYING to answer the questions because I'm so confused...

Homework Equations





The Attempt at a Solution

 
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OMG nevermind...a million hours later i looked at this again, and i have no idea what i was thinking earlier...

I figured it out...this is why it's good to get a good night's sleep...after a while, even simple things look impossible! hahahah
 
For those looking later at the thread - mass of the solution is NOT mass of the reagent dissolved.
 

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