Stoichiometry: Identify the metal and formula of it

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SUMMARY

The discussion focuses on identifying the lanthanide metal from a 0.250g sample reacted with excess hydrochloric acid (HCl), resulting in a product weight of 0.427g. The key conclusion is that the metal is Samarium (Sm), which forms the compound Samarium(III) chloride (SmCl3). The reaction can be simplified to Sm + 3HCl → SmCl3 + 1.5H2, where the hydrogen gas is not essential for solving the stoichiometric problem. The critical aspect of the equation is the stoichiometric ratio between Sm and SmCl3.

PREREQUISITES
  • Understanding of stoichiometry and chemical reactions
  • Familiarity with lanthanide series and their properties
  • Knowledge of balancing chemical equations
  • Basic principles of mass conservation in chemical reactions
NEXT STEPS
  • Study the properties and reactions of lanthanide metals
  • Learn about stoichiometric calculations in chemical reactions
  • Explore the synthesis and applications of Samarium(III) chloride
  • Research the role of hydrogen gas in acid-metal reactions
USEFUL FOR

Chemistry students, educators, and professionals involved in analytical chemistry or materials science, particularly those focusing on lanthanide elements and stoichiometric calculations.

Sho Kano
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Homework Statement


All of the lanthanide metals (La through Lu) react with HCL to form compounds having eithe the formula MCl2, MCl3, MCl4, (where M represents the metallic element). Each metal forms a single compound. A chemist has a 0.250g sample of a lanthanide metal, and she wishes to identify the metal. She reacts the metal with excess HCl and obtains 0.427g of the product. Based on this information, identify the metal and write the chemical formula of the product.

Homework Equations

The Attempt at a Solution


I'm not sure the below equation is correct. I originally wrote the equation without the H2, but it did not seem to be right with conservation of mass.
M + HCL —> MCl2 (or MCl3, MCl4) + H2
Basically, if the H2 does belong there, then I'd just have to guess MCl2, 3, or 4, then balance the equation, then randomly guess an element right? Seems like a lot of work.
 
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Sho Kano said:
and obtains 0.427g of the product.
Sho Kano said:
2, 3, or 4, then balance
 
I can't see anything on your post
 
MCln is all you need; ignore the H2, and just start cutting and trying.
 
Bystander said:
MCln is all you need; ignore the H2, and just start cutting and trying.
I'm getting Sm as the unknown metal; is that right?
 
Samarium? Mental arithmetic checks out.
 
Bystander said:
Samarium? Mental arithmetic checks out.
Awesome. What happens to the hydrogen though?
 
Sho Kano said:
What happens to the hydrogen though?
It "fizzes" away up the chimney/hood/exhaust system.
 
Bystander said:
It "fizzes" away up the chimney/hood/exhaust system.
I mean why is it not represented in the chemical equation
 
  • #10
"M" + nHCl = (n/2)H2 + "M"Cln; it is.
 
  • #11
Hydrogen is present in the reaction equation and we can even easily calculate its amount. We just ignore it as it is not necessary for solving the problem (just like the amount of HCl doesn't matter, as long as there is enough of it).

The only part of the equation that matters is

Sm → SmCl3

It preserves the only important information here - stoichiometric ratio between Sm and SmCl3. When solving this particular problem everything else is just a noise.
 

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