Calculating Mass in a Gas Stoichiometry Problem

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SUMMARY

The discussion focuses on a stoichiometry problem involving the reaction of zirconium metal with chlorine gas to form zirconium (IV) chloride, represented by the equation Zr + 2Cl2 = ZrCl4. The user successfully calculated that 400mL of chlorine gas is required to produce 200mL of ZrCl4 at 350 degrees Celsius and 50 kPa. For the mass of zirconium consumed, the user initially calculated 0.176 g using the moles of chlorine gas, confirming the accuracy of their approach after correcting a volume error in their calculations.

PREREQUISITES
  • Understanding of gas laws, specifically the ideal gas law (PV = nRT).
  • Knowledge of stoichiometry and mole ratios in chemical reactions.
  • Familiarity with calculating molar mass, particularly for zirconium (Zr).
  • Basic skills in unit conversion and dimensional analysis.
NEXT STEPS
  • Study the ideal gas law and its applications in stoichiometry problems.
  • Learn about mole-to-mole conversions in chemical reactions.
  • Explore the concept of limiting reactants in stoichiometric calculations.
  • Review molar mass calculations for various elements and compounds.
USEFUL FOR

Chemistry students, educators, and anyone preparing for exams involving stoichiometry and gas laws will benefit from this discussion.

Lili123
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Hi all,
While studying for my exam (and doing review questions) I came across a problem that I couldn't solve:

5. Zirconium metal and chlorine gas react to form zirconium (IV) chloride
Zr + 2Cl2 = ZrCl4
a) What volume of chlorine gas must be used at 350 degrees Celsius and 50 kPa to produce 200mL of ZrCl4 under the same conditions?

Figured out this part, and it comes to 400mL of chlorine gas.

b) What mass of zirconium will be used up?This part I had trouble on. I used the moles of chlorine gas to do this, though it feels like I'm doing something wrong, could someone let me know if what I am doing is right? We don't have an answer key... Thanks in advance!

0.003860124 mols Cl2 * 1 mol Zr / 2 mols Cl2 * 91.224g/mol Zr = 0.176 g Zr
 
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Lili123 said:
0.003860124 mols Cl2

Why 0.00386 moles of Cl2?
 
Borek said:
Why 0.00386 moles of Cl2?
Because in part A of the question I determined the moles of ZrCl4 using this formula: n = PV / RT and then used the mole ratio to convert to Cl2.
n = (50) (0.200) / (8.3145) (623.15) * 2mols Cl2 / 1mol ZrCl4 = 0.003860124
 
Oops, you are right. I used a wrong volume in my calculations.

0.176 g of Zr is a correct answer.
 

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