Calculating HCl Volume for Na2CO3 Reaction

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    Hcl Stoichiometry
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SUMMARY

The discussion centers on calculating the volume of 3.0 mol HCl needed to react with 10.0g of Na2CO3. Participants clarify that the term "volume" likely refers to a concentration of 3.0 M HCl, indicating an aqueous solution rather than a gas. The stoichiometric relationship between Na2CO3 and HCl is essential for determining the correct volume. The confusion arises from the initial interpretation of "volume" versus "moles" in the context of the reaction.

PREREQUISITES
  • Understanding of stoichiometry in chemical reactions
  • Knowledge of molarity (M) and its application in solution calculations
  • Familiarity with the properties of hydrochloric acid (HCl) and sodium carbonate (Na2CO3)
  • Basic principles of gas laws if considering HCl as a gas
NEXT STEPS
  • Calculate the molar mass of Na2CO3 to determine moles from grams
  • Learn how to convert moles of HCl to volume using molarity
  • Study the balanced chemical equation for the reaction between Na2CO3 and HCl
  • Explore the implications of using HCl in both aqueous and gaseous states
USEFUL FOR

Chemistry students, educators, and professionals involved in chemical reactions, particularly those focusing on acid-base reactions and solution concentrations.

Johnny_07
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Calculate the volume of 3.0 mol HCl required to react completely with 10.0g of Na2CO3.

Im confused about the "volume". What do they mean ?
Do I have to find the mass of HCl then convert it into a volume?
 
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Are you sure it is not 3.0 M of HCl?

If it's volume, check out density after finding the stoichiometric constant for the balance equation and finding the mol of sodium bicarbonate.
 
kingdomof said:
If it's volume, check out density after finding the stoichiometric constant for the balance equation and finding the mol of sodium bicarbonate.

If it is M (mol/L) you don't need density.
 
unless you are forgetting that HCl is a gas unless specified otherwise (aqueous solution or sth); so by volume they mean the volume of the required gas under normal/standard conditions
 
But we're pretty sure the OP meant 3.0 M HCl, which does imply a solution, not a gas.

If it were really 3.0 moles as stated, it would be pointless to talk about the amount "required to react completely with 10.0g of Na2CO3".
 

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